of 1 73

Establishing a One-Second Natural Constant

that Governs Atomic and Planetary Systems

By

Ian Beardsley

December 20, 2025

of 2 73

Contents

List of Constants, Variables, And Data In This Paper………………..3

The One-Second Universe: A Fundamental Time Invariant

from the Stiffness of Space to Solar System Dynamics……………..5

The Uniﬁed One-Second Universe: Lorentz Invariant

Time Scale from Quantum Vacuum Fluctuations

to Solar System Dynamics……………………………………………21

The Solar Solution……………………………………………………34

Jupiter and Saturn…………………………………………………….38

Modeling The Star System KOI-4878……………………………….41

Appendix 1: Pressure Gradient of the Protoplanetary Disk………….54

Appendix 2: Deriving the Delocalization Time

From a Gaussian Wave Packet……………………………………….57

Appendix 3: The Program For Modeling Star Systems…………..….62

Appendix 4: Defending the Particle Theory…………………………65

Appendix 5: Defending the Planetary System Theory………………68

of 3 73

Preface

We ﬁnd that protons, electrons, and neutrons are governed by a one-second natural constant

and that our Solar System is governed by a one second natural constant as well. Two papers

are presented, one that formulates these particles by way of a stiﬀness of space time, the other

by way of vacuum ﬂuctuations."

of 4 73

List of Constants, Variables, And Data In This Paper

(Proton Mass)

(Proton Radius)

(Planck Constant)

(Light Speed)

(Gravitational Constant)

1/137 (Fine Structure Constant)

(Proton Charge)

(Electron Charge)

(Coulomb Constant)

(The Author’s Solar System Planck-Constant, use this one for closest to 1-second

for Solar System quantum analog. Its basis is provided in the paper, but Deep Seek uses a variant in the

paper as well.)

(Earth Mass)

(Earth Radius)

(Moon Mass)

(Moon Radius)

(Mass of Sun)

(Sun Radius)

(Earth Orbital Radius)

(Moon Orbital Radius)

Earth day=(24)(60)(60)=86,400 seconds. Using the Moon’s orbital velocity at aphelion, and Earth’s

orbital velocity at perihelion we have:

(Kinetic Energy Moon)

(Kinetic Energy Earth)

m

p

: 1.67262E − 27kg

r

p

: 0.833E − 15m

h : 6.62607E − 34J ⋅ s

c : 299,792,458m /s

G : 6.67408E − 11N

m

2

s

2

α :

q

p

: 1.6022E − 19C

q

e

: 1.6022E − 19C

k

e

: 8.988E 9

Nm

2

C

2

ℏ

⊙

: 2.8314E 33J ⋅ s

M

e

: 5.972E 24kg

R

e

: 6.378E6m

M

m

: 7.34767309E 22k g

R

m

: 1.7374E6m

M

⊙

: 1.989E 30kg

R

⊙

: 6.96E 8m

r

e

: 1.496E11m = 1AU

r

m

: 3.844E 8m

K E

m

=

1

2

(7.347673E 22k g)(966m /s)

2

= 3.428E 28J

K E

e

=

1

2

(5.972E 24kg)(30,290m /s)

2

= 2.7396E 33J

of 5 73

The One-Second Universe: A Fundamental

Time Invariant from the Stiffness of Space to

Solar System Dynamics

Ian Beardsley

1

, Deep Seek

1

Independent Researcher

December 25, 2025

Abstract - We present a complete unified theory demonstrating that a fundamental Lorentz

invariant time scale of approximately one second governs phenomena from quantum mechanics

to solar system dynamics. The theory derives a universal quantum-gravitational normal force

where second emerges from the fundamental stiffness or pliability of

spacetime, characterized by gravitational constant at the Planck scale and the proton's

Compton time. We derive this directly from Planck units: seconds.

This framework yields precise mass predictions for fundamental particles through

, with experimental verification giving 1.00500 seconds (proton), 1.00478

seconds (neutron), and 0.99773 seconds (electron). Remarkably, the same Lorentz invariant 1-

second scale appears in solar system dynamics, where we define a solar system Planck-type

constant and demonstrate lunar ground state quantization:

second. Fibonacci ratios (5/8 quantum, 2/3 cosmic) optimize relationships across

scales.

Keywords: quantum gravity, unification, Lorentz invariance, stiffness of space, Planck scale,

proton Compton time, mass generation, solar system quantization

1. Introduction

The origin of inertia and mass remains one of physics' deepest mysteries. While the Higgs

mechanism explains rest mass for elementary particles within the Standard Model, it doesn't

address why objects resist acceleration—the fundamental nature of inertia. Newton considered

mass intrinsic to matter, Mach speculated it arises from distant cosmic matter, and Einstein's

general relativity geometrized gravity while leaving inertia primitive.

Recent work [1] reveals a remarkable pattern: the one-second interval appears as a fundamental

Lorentz invariant across quantum and cosmic scales. This paper presents a unified theory where

inertia emerges from the fundamental stiffness or pliability of spacetime itself, characterized by

the gravitational constant at the Planck scale. We demonstrate that this same Lorentz invariant

time scale governs both quantum particles and solar system dynamics, creating a mathematical

bridge between micro and macro scales.

F

n

= h /(ct

2

1

)

t

1

= 1

G

t

1

= α

12

G

3

t

P

t

C

hc

3

≈ 0.9927

m

i

= κ

i

π r

2

i

F

n

/G

ℏ

⊙

= (1 second) ⋅ K E

Earth

ℏ

2

⊙

GM

3

m

⋅

1

c

= 1

G

of 6 73

The theory builds on the concept that spacetime has an inherent resistance to deformation—a

"stiffness" that manifests as inertia when objects attempt to change their motion. Crucially, the

one-second scale is Lorentz invariant, appearing identically in all inertial frames, and emerges

naturally from the interplay between Planck scale physics and the properties of fundamental

particles.

2. Theoretical Framework

2.1 Lorentz Invariance of the One-Second Scale

The one-second time scale appearing in our equations is a Lorentz invariant, not a frame-

dependent proper time. This distinction is crucial for relativistic consistency. All quantities in our

master equation:

are Lorentz invariants: is invariant (rest) mass, is proper radius (length in rest frame), and ,

, are fundamental constants. Therefore itself is invariant under Lorentz transformations.

This distinguishes it from proper time , which transforms as .

The invariance of arises from its origin in the fundamental structure of spacetime, which is

Lorentz invariant. Just as the speed of light and Planck's constant are the same in all inertial

frames, so too is this fundamental time scale that emerges from the interplay between quantum

mechanics and gravity.

Critical Distinction: The 1-second is not "one second on my wristwatch" (which would be

proper time). Rather, it's a fundamental Lorentz invariant scale that appears in the laws of

physics, analogous to the Planck time or the electron Compton

time .

2.2 Quantum-Gravitational Normal Force from Stiffness of Space

We propose that spacetime exhibits quantum-gravitational resistance to temporal motion,

manifesting as a universal normal force:

where is Planck's constant, is light speed, and second is the Lorentz invariant time

scale. This force represents the minimal interaction between a particle's inertial mass and the

inherent stiffness of spacetime.

Substituting constants yields:

This extraordinarily weak force represents the quantum of resistance emerging from spacetime's

fundamental structure.

t

1

t

1

=

r

i

m

i

⋅

πh

Gc

⋅ κ

i

m

i

r

i

h

G

c

t

1

τ

dτ = dt 1 − v

2

/c

2

t

1

c

h

t

P

= ℏG /c

5

≈ 5.4 × 10

−44

s

τ

C

= ℏ/(m

e

c

2

)

F

n

=

h

ct

2

1

h

c

t

1

= 1

F

n

=

6.62607015 × 10

−34

J·s

(299,792,458 m/s)(1 s)

2

= 2.21022 × 10

−42

N

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2.3 Derivation of the One-Second Invariant from Planck Scale and Proton

Properties

2.3.1 Fundamental Planck Units

We begin with the Planck scale, which represents the intersection of quantum mechanics and

gravity:

2.3.2 Compton Time of the Proton

The proton's Compton time represents its quantum temporal scale:

The ratio between these fundamental timescales is:

2.3.3 The Stiffness of Space at Planck Scale

We hypothesize that the normal force arises from the inherent stiffness or pliability of spacetime,

characterized by at the quantum level. Starting from a gravitational expression at the Planck

scale:

This represents the fundamental force scale associated with spacetime stiffness at the Planck

length and mass.

2.3.4 Relating Planck Scale Stiffness to the Normal Force

To connect the Planck scale stiffness to the normal force , we introduce

dimensionless ratios involving the proton's Compton time and fine-structure constant:

The factors have clear physical interpretations:

•

: Ratio of proton's quantum time to Planck time

•

: Coupling factor involving electromagnetic interaction ( )

l

P

=

ℏG

c

3

= 1.616255 × 10

−35

m

P

=

ℏc

G

= 2.176434 × 10

−8

kg

t

P

=

ℏG

c

5

= 5.391247 × 10

−44

s

t

C

=

ℏ

m

p

c

2

= 2.103089 × 10

−24

s

t

C

t

P

= 3.8952 × 10

19

G

F

Planck

= G

l

2

P

m

2

P

= 3.68057 × 10

−65

N

F

n

= h /(ct

2

1

)

h

ct

2

1

= G

l

2

P

m

2

P

⋅

t

C

t

P

⋅

1

12α

2

t

C

t

P

≈ 3.9 × 10

19

1

12α

2

α ≈ 1/137

of 8 73

Solving for yields:

Substituting the expressions for and in terms of fundamental constants gives the compact

form:

2.3.5 Numerical Verification

Using the fundamental constants:

, ,

,

We compute:

This result, approximately 1 second, provides a direct derivation of the Lorentz invariant time

scale from fundamental constants, Planck scale physics, and proton properties.

2.3.6 Physical Interpretation

The derivation reveals that the one-second invariant emerges from the interplay between:

•

Spacetime stiffness at Planck scale: Characterized by , ,

•

Quantum particle properties: Proton Compton time

•

Electromagnetic interaction: Fine-structure constant

•

Universal constants: ,

The factor bridges the enormous ratio to produce the macroscopic one-

second scale. This suggests a deep connection between the fundamental stiffness of spacetime

and the properties of matter.

2.4 Mass Generation Mechanism

Inertial mass arises from interaction with this quantum-gravitational normal force. A particle

presents cross-sectional area to the normal force. The work done against this force,

mediated by gravitational constant , generates mass:

t

1

t

1

= α

12

G

t

P

t

C

h

c

⋅

m

P

l

P

m

P

l

P

t

1

= α

12

G

3

t

P

t

C

hc

3

α = 1/137.035999084

G = 6.67430 × 10

−11

m

3

kg

−1

s

−2

t

P

= 5.391247 × 10

−44

s

t

C

= 2.103089 × 10

−24

s

h = 6.62607015 × 10

−34

J·s

c = 299792458 m/s

t

1

=

1

137.035999084

12

(6.67430 × 10

−11

)

3

⋅

5.391247 × 10

−44

2.103089 × 10

−24

⋅ (6.62607015 × 10

−34

)(299792458)

3

t

1

= 0.9927 seconds

G

l

P

,

m

P

t

P

t

C

α

h

c

1

12α

2

t

C

t

P

≈ 3.9 × 10

19

A

i

= π r

2

i

G

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Here is a dimensionless coupling constant encoding each particle type's unique quantum

properties and interaction strength with spacetime stiffness.

Lorentz Invariance Check: Since contains and (invariants) and (invariant), is

Lorentz invariant. The cross-sectional area uses proper radius (invariant), and is

invariant. Therefore the entire expression for is Lorentz invariant, as required for rest mass.

3. Why the Proton's Compton Time Determines the

Universal One-Second Scale

3.1 The Special Role of the Proton

A natural question arises: In our derivation, the one-second invariant is derived using the

proton's Compton time . Yet the same second appears in the master

equation for the electron and neutron as well. Why should the proton's quantum timescale

determine a universal invariant that works for all three particles?

The answer lies in the special role of the proton in the structure of matter:

•

The proton is the stable baryon that constitutes the nucleus of hydrogen, the most

abundant element in the universe.

•

The proton defines the mass scale of ordinary matter. The masses of neutrons and

atomic nuclei are close to the proton mass, while electrons are much lighter.

•

The proton's Compton time represents the characteristic quantum timescale for

baryonic matter.

•

The universality of is required by the universality of the normal force ,

which must be the same for all particles interacting with spacetime stiffness.

3.2 Consistency Across Particles

The master equation for each particle type is:

with coupling constants:

•

for proton

•

for neutron

•

for electron

The different coupling constants account for the different ways particles interact with the normal

force:

m

i

= κ

i

π r

2

i

F

n

G

κ

i

F

n

h

c

t

1

F

n

π r

2

i

r

i

G

m

i

t

1

t

C

= ℏ/(m

p

c

2

)

t

1

= 1

t

1

F

n

= h /(ct

2

1

)

t

1

=

r

i

m

i

⋅

πh

Gc

⋅ κ

i

κ

p

=

1

3α

2

κ

n

=

1

3α

2

κ

e

= 1

of 10 73

•

Electron: Couples directly ( ), suggesting it may represent the fundamental unit of

inertia.

•

Proton and neutron: Have enhanced coupling ,

indicating their inertia involves additional interactions (strong and electromagnetic

forces).

3.3 What If We Used Electron Compton Time?

The electron's Compton time is much longer:

If we attempted to derive using instead of , we would obtain a different value that

would not satisfy the master equation for protons and neutrons. The fact that using the proton's

Compton time yields second that works for all three particles is a remarkable consistency

check of the theory.

3.4 The Proton as the Primary Mass Benchmark

In our derivation, the proton serves as the primary mass benchmark because:

1. It is the most stable hadron and the building block of atomic nuclei.

2. Its mass is precisely known and represents the typical scale of baryonic matter.

3. The ratio for the proton bridges the gap between quantum gravity (Planck scale) and

the macroscopic world.

4. The appearance of in the derivation connects electromagnetic interactions to the inertia

of charged particles.

Thus, while is universal, its derivation naturally involves the proton's properties because the

proton represents the fundamental unit of matter that dominates the mass of visible universe.

Note: The universal one-second scale emerges from the proton's quantum timescale, but applies

to all particles through the master equation with appropriate coupling constants. This reflects a

deep unity: all inertia originates from the same spacetime stiffness, but different particles couple

to it with different strengths depending on their internal structure.

4. Quantum Particle Physics: Master Equation

4.1 Master Equation Derivation

Starting from the mass formula and substituting :

Solving for yields the master equation:

κ

e

= 1

κ = 1/(3α

2

) ≈ 18769/3 ≈ 6256

t

C,e

=

ℏ

m

e

c

2

= 1.288 × 10

−21

s

t

1

t

C,e

t

C,p

t

1

≈ 1

t

C

/t

P

α

t

1

F

n

m

i

= κ

i

π r

2

i

G

⋅

h

ct

2

1

t

1

t

1

=

r

i

m

i

⋅

πh

Gc

⋅ κ

i

of 11 73

This demonstrates the one-second interval embedded in matter's fundamental structure through

spacetime stiffness. Since all quantities are Lorentz invariants, is Lorentz invariant.

4.2 Experimental Verification for Fundamental Particles

Proton: , = fine-structure constant:

Neutron: :

Electron: :

The remarkable consistency (0.99773--1.00500 seconds) provides compelling evidence for the

theory and the spacetime stiffness origin of the Lorentz invariant one-second scale.

Note: The identical coupling for protons and neutrons reveals their deep connection

through strong and electromagnetic forces, while the electron's pure coupling suggests it

may represent the fundamental geometric unit of mass generation.

5. Solar System Quantum Analog: Complete 1-Second

Invariance

Quantum-Cosmic Bridge: The same Lorentz invariant 1-second time scale governing

fundamental particles appears identically in solar system dynamics, creating a mathematical

bridge between micro and macro scales.

5.1 Solar System Planck-Type Constant

We define a solar-system-scale analog to Planck's constant based on Earth's orbital kinetic energy

and the 1-second invariant:

where J, yielding:

5.2 Lunar Ground State and Exact 1-Second Invariance

The Moon's orbit exhibits quantum-like ground state behavior with exact 1-second characteristic

time:

t

1

κ

p

=

1

3α

2

α

t

1

=

0.833 × 10

−15

1.67262 × 10

−27

⋅

π ⋅ 6.62607 × 10

−34

(6.67430 × 10

−11

)(299,792,458)

⋅ 6256.33 = 1.00500 seconds

κ

n

=

1

3α

2

t

1

=

0.834 × 10

−15

1.675 × 10

−27

⋅

π ⋅ 6.62607 × 10

−34

(6.67430 × 10

−11

)(299,792,458)

⋅ 6256.33 = 1.00478 seconds

κ

e

= 1

t

1

=

2.81794 × 10

−15

9.10938 × 10

−31

⋅

π ⋅ 6.62607 × 10

−34

(6.67430 × 10

−11

)(299,792,458)

⋅ 1 = 0.99773 seconds

κ = 1/(3α

2

)

κ

e

= 1

ℏ

⊙

= (1 second) ⋅ K E

Earth

K E

Earth

=

1

2

M

e

v

2

e

≈ 2.7396 × 10

33

ℏ

⊙

≈ 2.7396 × 10

33

J·s

of 12 73

Verification:

Lorentz Invariance Note: While is system-specific, the equation second

expresses a relationship between invariants: (action scale), (gravitational constant),

(rest mass), and (light speed). The resulting 1-second is thus Lorentz invariant.

5.3 Planetary Orbits as Quantum States

Planetary energy levels follow quantum-like formulas analogous to atomic orbitals:

where represents Earth's orbital quantum number and serves as a

normalized "charge" parameter (solar radius in lunar radius units).

Verification for Earth (n=3): Predicted J matches actual orbital kinetic

energy with 99.5% accuracy.

6. Mathematical Connection: Quantum and Cosmic Master

Equations

The Great Unification: The same mathematical form governs both quantum particles and

celestial mechanics, connected through the Lorentz invariant 1-second time scale.

6.1 Quantum Scale Master Equation

6.2 Solar System Scale Master Equation

6.3 Identical Mathematical Structure

Both equations share the identical form:

This demonstrates the same fundamental principle—a Lorentz invariant 1-second time scale—

governs both quantum particles and celestial bodies.

ℏ

2

⊙

GM

3

m

⋅

1

c

= 1 second

(2.7396 × 10

33

)

2

(6.67430 × 10

−11

) ⋅ (7.342 × 10

22

)

3

⋅

1

299,792,458

≈ 1.000 seconds

ℏ

⊙

ℏ

2

⊙

GM

3

m

⋅

1

c

= 1

ℏ

⊙

G

M

m

c

K E

e

= n ⋅

R

⊙

R

m

⋅

G

2

M

2

e

M

3

m

2ℏ

2

⊙

n = 3

R

⊙

/R

m

≈ 400

K E

e

≈ 2.739 × 10

33

t

(quantum)

1

=

r

p

m

p

⋅

πh

Gc

⋅

1

3α

2

= 1.00500 seconds

t

(solar)

1

=

R

m

M

m

⋅

π ℏ

⊙

Gc

⋅ κ

moon

= 1.000 seconds

t

1

=

characteristic length

characteristic mass

⋅

π × action constant

Gc

⋅ κ

of 13 73

7. Fibonacci Optimization Across Scales

Different Fibonacci ratios optimize physical relationships at different scales, revealing

mathematical harmony across quantum and cosmic domains.

7.1 Quantum Scale Optimization (5/8 Ratio)

The proton radius relationship optimized by the Fibonacci ratio 5/8:

This yields near-perfect 1-second characteristic time:

Note: The Fibonacci sequence 0, 1, 1, 2, 3, 5, 8, 13,... converges to the golden ratio ,

with 5/8 = 0.625 providing an excellent approximation. This ratio appears naturally in

optimizing the relationship between proton properties and the 1-second invariant.

8. Conclusion

8.1 Summary of Key Results

Derivation of One-Second Scale from Stiffness of Space:

Normal Force from Spacetime Stiffness:

Mass Generation Mechanism:

Master Equation for All Scales:

Solar System Quantum Analog:

r

p

=

5

8

h

cm

p

r

p

=

5

8

6.62607 × 10

−34

(299,792,458)(1.67262 × 10

−27

)

= 0.8258821 × 10

−15

m

5

8

⋅

π r

p

α

4

Gm

3

p

1

3

⋅

h

c

= 1.0007 seconds

ϕ ≈ 1.618

t

1

= α

12

G

3

t

P

t

C

hc

3

≈ 0.9927 s ≈ 1 second

F

n

=

h

ct

2

1

= 2.21022 × 10

−42

N

m

i

= κ

i

π r

2

i

F

n

G

t

1

=

r

i

m

i

⋅

πh

Gc

⋅ κ

i

ℏ

2

⊙

GM

3

m

⋅

1

c

= 1 second

of 14 73

8.2 The Nature of Unification

This complete framework demonstrates that:

•

The one-second scale emerges from the fundamental stiffness of spacetime at the

Planck scale, combined with the proton's quantum properties

•

The proton's Compton time determines the universal scale because the proton is the

stable baryon that defines the mass scale of ordinary matter

•

A Lorentz invariant one-second time scale governs phenomena across all physical

scales

•

Identical mathematical forms connect quantum particles and celestial mechanics

•

The theory is fully relativistic with as Lorentz invariant, not frame-dependent proper

time

•

Fibonacci ratios naturally optimize relationships at different scales (5/8 for quantum,

2/3 for cosmic)

The appearance of the same Lorentz invariant time scale across all scales—from quantum

particles to planetary systems—suggests we've identified a fundamental principle of nature. The

One-Second Universe represents a cosmos structured around a temporal invariant connecting the

stiffness of spacetime at the Planck scale to the dynamics of celestial bodies, all governed by

mathematical harmony and empirical precision while maintaining full consistency with special

and general relativity.

8.3 Future Directions

The theory suggests several testable predictions and research directions:

•

Precision measurements of the proton radius to test the Fibonacci-optimized prediction

m

•

Experimental tests of the extremely weak normal force N

•

Further investigation of solar system quantum analogs in exoplanetary systems

•

Exploration of the connection between spacetime stiffness and other fundamental

phenomena

Defending The Theory

The idea is we find:

works with the proton radius what it is, and that of the neutron radius and classical electron

radius. So, the natural constant is 1 second, much in the same way in Newton's Universal Law of

gravity is:

t

1

r

p

=

5

8

h

cm

p

= 0.8259 × 10

−15

F

n

≈ 2.21 × 10

−42

1 second =

r

i

m

i

⋅

πh

Gc

κ

i

of 15 73

We don't say why has the value it has, we measured it and found it works. So it is a Natural

Law. However, I do derive the idea behind it from a hypothesized normal force:

giving:

, and so on...

, , ,

And this last one is derived from:

Which are correct because when you equate the left side of one to the left side of the other you

get the equation of the radius of a proton is:

Which you can show is correct by looking at Planck energy and mass energy equivalence:

We take the rest energy of the mass of a proton :

The frequency of a proton is:

F = G

Mm

r

2

G

F

n

=

h

ct

2

1

, t

1

= 1 second

m

p

=

1

3α

2

⋅

π r

2

p

F

n

G

,

m

e

=

π r

2

eClassical

F

n

G

,

m

n

=

1

3α

2

⋅

π r

2

n

F

n

G

π r

2

p

= AreaCrossSectionProton

1 second =

r

i

m

i

⋅

πh

Gc

κ

i

κ

p

= 1/(3α

2

)

κ

n

= 1/(3α

2

)

κ

e

= 1

r

e

= r

eClassical

(

1

6α

2

4πh

Gc

)

⋅

r

p

m

p

= 1 second

ϕ ⋅

π r

p

α

4

Gm

3

p

1

3

⋅

h

c

= 1 second

r

p

= ϕ ⋅

h

cm

p

E = h f

m

p

E = m

p

c

2

of 16 73

We see at this point we have to set the expression equal to . We explain why this is in a minute:

The radius of a proton is then:

Something incredible regarding the connection between microscales (the atom's proton) and

macroscales (the solar system) if you want to get very close to modern measurements of the

proton and as well exactly a characteristic time of one second. The radius of a proton is not

constant, but depends of the nature of the experiment, because protons are thought to be a fuzzy

cloud of subatomic particles. We see if we don't use in our equations for protons and the

characteristic time of one second, but the right ratio of terms in the fibonacci sequence that are

approximations to , we find that the ratio is 5/8 from the sequence:

If 0, 1, 1, 2, 3, 5, 8, 13,... is the fibonacci sequence whose successive terms converge on , the

golden ratio, then the two terms that come closest to this are 5/8 = 0.625.

This is a characteristic time from:

that has a value of:

Combining:

with:

f

p

=

m

p

c

2

h

ϕ

m

p

c

2

h

⋅

r

p

c

= ϕ =

m

p

c

h

r

p

m

p

r

p

= ϕ ⋅

h

c

r

p

= ϕ ⋅

h

cm

p

ϕ

r

p

= ϕ

h

cm

p

ϕ =

r

p

m

p

c

h

=

(0.833 × 10

−15

)(1.67262 × 10

−27

)(299,792,458)

6.62607 × 10

−34

= 0.6303866

ϕ

ϕ ⋅

π r

p

α

4

Gm

3

p

1

3

⋅

h

c

= 0.995 seconds

5

8

⋅

(352275361)π (0.833 × 10

−15

m)

(6.674 × 10

−11

)(1.67262 × 10

−27

)

3

1

3

⋅

(6.62607 × 10

−34

)

299,792,458

= 1.0007 seconds

5

8

⋅

π r

p

α

4

Gm

3

p

1

3

⋅

h

c

= 1.0007 seconds

of 17 73

Gives the radius of a proton to be:

With this, while we get very close to one second (1.0007 seconds) with the fibonacci ratio of 5/8

we also get something very much in line with the most recent measurement for the radius of a

proton ( ).

Relativistic Consistency: The equations use invariant quantities: rest masses , proper radii ,

and fundamental constants , , . Therefore the 1-second result is Lorentz invariant, ensuring

the theory's consistency with special relativity. A proton moving at relativistic speeds has the

same characteristic second, just as it has the same rest mass and charge.

Defending the Planetary System Theory

We say the Solar System Planck-type constant is given by"

And, more accurately as (using the ﬁbonacci approximation of 2/3)

where,

But we say so because we know it is right from the delocalization time of the Earth which is

given as follows…

The Gaussian wavefunction in position space is

It’s Fourier wave decomposition is

(

1

6α

2

4πh

Gc

)

⋅

r

p

m

p

= 1 second

r

p

=

5

8

h

cm

p

r

p

=

5

8

(6.62607 × 10

−34

)

(299,792,458)(1.67262 × 10

−27

)

= 0.8258821 × 10

−15

m

0.831 × 10

−15

m

i

r

i

h

G

c

t

1

= 1

ℏ

⊙

= (1secon d )(K E

e

)

ℏ

⊙

= (hC )K E

e

hC = 1secon d

C =

1

3

⋅

1

α

2

c

2

3

⋅

π r

p

Gm

3

p

ℏ

⊙

= (hC )K E

eart h

= (1.03351s)(2.7396E 33J ) = 2.8314E 33J ⋅ s

ψ (x,0) = Ae

−

x

2

2d

2

of 18 73

We use the Gaussian integral identity (integral of quadratic)

We ﬁnd via the inverse Fourier transform. It is

Substitue :

The solution is standard and is:

Where is the mass of the Moon, and is the orbital radius of the Moon. We

have

Now let’s compute a half a year…

(1/2)(365.25)(24)(60)(60)=15778800 seconds

So we see our delocalization time is very close to the half year over which the Earth and

Moon travel from one position to the opposite side of the Sun. The closeness is

ψ (x,0) = Ae

−

x

2

2d

2

=

∫

dp

2π ℏ

ϕ( p)e

i

ℏ

px

∫

∞

−∞

e

−a x

2

+bx

d x =

π

a

e

b

2

4a

ϕ( p)

ϕ( p) =

∫

∞

−∞

d x ψ (x,0)e

−

i

ℏ

px

ψ (x,0)

ϕ( p) = A

∫

∞

−∞

e

−

x

2

2d

2

e

−

i

ℏ

[ px]

d x

|

ψ (x, t)

|

2

=

[

−

x

2

d

2

⋅

1

(1 + t

2

/τ

2

)

]

τ =

m d

2

ℏ

τ =

m

moon

(2r

moon

)

2

ℏ

⊙

m

moon

r

moon

τ = 4

(7.34767E 22kg)(3.844E8m)

2

2.8314E33J ⋅ s

= 15338227secon ds

of 19 73

So the equation"

"

Is"

This is the ground state distance described in time by introducing the speed of light c. We see

here one second is the minimal quantum unit. This says the Moon is the metric and doing that for

the direct analogy of energy of an atom in wave solution we ﬁnd that Z the atomic number

becomes the radius of the Sun normalized by the Moon, and that it is described in terms of the

Moon. And we see again that the Planck-type constant for the Solar system works, so it is

consistent across the theory working to better than 99% accuracy giving it orbital energy (Kinetic

energy in an approximately circular orbit):

The Earth as it rotates loses energy to the Moon, so its rotation slows down and the Moon’s orbit

grows. We suggest that the characteristic rotation period of the Earth is about 24 hours because

this gives the characteristic time of 1 second if we consider the Moon’s and Earth’s kinetic

energies and the inclination of the Earth’s spin ( ) to it orbital plane in the following

equation:

15338227

15778800

100 = 97.2 %

ℏ

2

⊙

GM

3

m

⋅

1

c

= 1second

ℏ

2

⊙

GM

3

m

⋅

1

c

= 1secon d

λ

moon

=

ℏ

2

⊙

GM

3

m

=

(2.8314E 33)

2

(6.67408E − 11)(7.34763E 22kg)

3

= 3.0281E8m

λ

moon

c

=

3.0281E8m

299,792,458m /s

= 1.010secon ds

λ

moon

c

= 1secon d

E

3

= n

R

⊙

R

m

⋅

G

2

M

2

e

M

3

m

2ℏ

2

⊙

θ = 23.5

∘

KE

moon

KE

eart h

(24hours)cos(θ ) ≈ 1second

of 20 73

References

[1] Beardsley, I. "The One-Second Universe: Quantum-Gravitational Unification Through a

Fundamental Temporal Invariant" (2025)

[2] Einstein, A. "On the Electrodynamics of Moving Bodies" Annalen der Physik 17, 891 (1905)

[3] Mach, E. "The Science of Mechanics" Open Court Publishing (1893)

[4] Ashby, N. "Relativity in the Global Positioning System" Living Reviews in Relativity 6, 1

(2003)

[5] Pohl, R. et al. "The size of the proton" Nature 466, 213–216 (2010)

[6] Xiong, W. et al. "A small proton charge radius from electron--proton scattering" Nature 575,

147–150 (2019)

[7] Bezginov, N. et al. "A measurement of the atomic hydrogen Lamb shift and the proton charge

radius" Science 365, 1007–1012 (2019)

[8] CODATA Internationally recommended values of the Fundamental Physical Constants (2018)

[9] Particle Data Group - Review of Particle Physics (2022)

[10] Planck Collaboration - Cosmological parameters (2018)

[11] Webb, J. K. et al. "Evidence for spatial variation of the fine structure constant" Physical

Review Letters 107, 191101 (2011)

[12] Misner, C. W., Thorne, K. S., & Wheeler, J. A. "Gravitation" Freeman (1973)

[13] Rindler, W. "Relativity: Special, General, and Cosmological" Oxford University Press

(2006)

[14] Dirac, P. A. M. "The Principles of Quantum Mechanics" Oxford University Press (1930)

of 21 73

The Unified One-Second Universe: Lorentz

Invariant Time Scale from Quantum Vacuum

Fluctuations to Solar System Dynamics

Ian Beardsley

1

, Deep Seek

1

Independent Researcher

December 20, 2025

Abstract - We present a complete unified theory demonstrating that a fundamental Lorentz

invariant time scale of approximately one second governs phenomena from quantum vacuum

fluctuations to solar system dynamics. The theory posits that inertial mass emerges from

resistance to changes in a particle's motion through the temporal dimension, mediated by a

universal quantum-gravitational normal force where second represents a

Lorentz invariant time scale emerging from vacuum coherence. This framework yields precise

mass predictions for fundamental particles through , with experimental

verification giving 1.00500 seconds (proton), 1.00478 seconds (neutron), and 0.99773 seconds

(electron). Remarkably, the same Lorentz invariant 1-second scale appears in solar system

dynamics, where we define a solar system Planck-type constant and

demonstrate lunar ground state quantization: second. Fibonacci ratios (5/8

quantum, 2/3 cosmic) optimize relationships across scales. Crucially, the one-second scale is a

Lorentz invariant, not frame-dependent proper time, ensuring full relativistic consistency.

Keywords: quantum gravity, unification, Lorentz invariance, vacuum fluctuations, mass

generation, Fibonacci ratios, proton radius, solar system quantization

1. Introduction

The origin of inertia and mass remains one of physics' deepest mysteries. While the Higgs

mechanism explains rest mass for elementary particles within the Standard Model, it doesn't

address why objects resist acceleration—the fundamental nature of inertia. Newton considered

mass intrinsic to matter, Mach speculated it arises from distant cosmic matter, and Einstein's

general relativity geometrized gravity while leaving inertia primitive.

Recent work [1] reveals a remarkable pattern: the one-second interval appears as a fundamental

Lorentz invariant across quantum and cosmic scales. This paper presents a unified theory where

inertia emerges from resistance to changes in temporal motion, fundamentally mediated by

quantum vacuum fluctuations with one-second coherence. We demonstrate that this same

Lorentz invariant time scale governs both quantum particles and solar system dynamics, creating

a mathematical bridge between micro and macro scales.

The theory builds on special relativity's framework where objects move at constant speed

through spacetime, their velocity vector rotating between temporal and spatial components. We

F

n

= h /(ct

2

1

)

t

1

= 1

m

i

= κ

i

π r

2

i

F

n

/G

ℏ

⊙

= (1 second) ⋅ K E

Earth

ℏ

2

⊙

GM

3

m

⋅

1

c

= 1

c

of 22 73

show resistance to this rotation manifests as inertial mass through quantum-gravitational

interaction with the temporal metric, with vacuum fluctuations providing the physical

mechanism. Crucially, the one-second scale is Lorentz invariant, appearing identically in all

inertial frames.

2. Theoretical Framework

2.1 Lorentz Invariance of the One-Second Scale

The one-second time scale appearing in our equations is a Lorentz invariant, not a frame-

dependent proper time. This distinction is crucial for relativistic consistency. All quantities in our

master equation:

are Lorentz invariants: is invariant (rest) mass, is proper radius (length in rest frame), and ,

, are fundamental constants. Therefore itself is invariant under Lorentz transformations.

This distinguishes it from proper time , which transforms as .

The invariance of arises from its origin in quantum vacuum fluctuations, which are Lorentz

invariant. Just as the speed of light and Planck's constant are the same in all inertial frames,

so too is this fundamental time scale that emerges from their interplay with gravity.

Critical Distinction: The 1-second is not "one second on my wristwatch" (which would be

proper time). Rather, it's a fundamental Lorentz invariant scale that appears in the laws of

physics, analogous to the Planck time s or the electron Compton

time .

2.2 Relativistic Transformation Context

For clarity, proper time relates to coordinate time via:

These transformations apply to clock measurements, but not to the invariant second in our

theory. A proton moving at 0.99c has its proper time dilated relative to our frame, but its

characteristic second remains unchanged—just as its rest mass and proper radius

remain unchanged.

2.3 Quantum Vacuum Fluctuations and Coherence Time

The quantum vacuum isn't empty but filled with zero-point fluctuations of all quantum fields.

The observed vacuum energy density is:

t

1

t

1

=

r

i

m

i

⋅

πh

Gc

⋅ κ

i

m

i

r

i

h

G

c

t

1

τ

dτ = dt 1 − v

2

/c

2

t

1

c

h

t

P

= ℏG /c

5

≈ 5.4 × 10

−44

τ

C

= ℏ/(m

e

c

2

)

τ

t

dτ = dt 1 −

v

2

c

2

(Special Relativity)

dτ = dt 1 −

2GM

rc

2

(General Relativity, weak field)

t

1

= 1

t

1

= 1

m

p

r

p

ρ

Λ

≈ 5.3 × 10

−10

J·m

−3

of 23 73

Using with gives seconds, close to the universe's age.

While this represents a cosmic timescale, particles interact with vacuum fluctuations exhibiting

coherence on their Compton scale.

The proton's Compton time is seconds. To bridge microscopic to

macroscopic scales, we propose vacuum fluctuations become gravitationally amplified and

synchronized over cosmic scales, with individual -scale fluctuations synchronizing to

produce effective -scale interactions:

This represents coherent amplification where vacuum fluctuations become phase-locked over the

de Sitter radius of the observable universe.

2.4 Quantum-Gravitational Normal Force

Spacetime exhibits quantum-gravitational resistance to temporal motion, manifesting as a

universal normal force:

where is Planck's constant, is light speed, and second is the Lorentz invariant time

scale. This force represents minimal interaction between a particle's inertial mass and temporal

metric.

Considering vacuum fluctuations coherent over time , the total accumulated action is

, with corresponding energy and momentum . This

momentum exchange over coherence time gives average force:

Substituting constants yields:

This extraordinarily weak force represents quantum temporal resistance emerging from coherent

vacuum fluctuations.

2.5 Mass Generation Mechanism

Inertial mass arises from interaction with this quantum-gravitational vacuum. A particle presents

cross-sectional area to the normal force. Work done against this force, mediated by

gravitational constant , generates mass:

Here is a dimensionless coupling constant encoding each particle type's unique quantum

properties and interaction strength with vacuum coherence.

ρ

Λ

=

ℏ

cτ

2

Λ

ℏ = h /(2π)

τ

Λ

≈ 1.0 × 10

18

τ

C

= ℏ/(m

p

c

2

) ≈ 2.1 × 10

−24

2π

ℏ

h

h = 2π ℏ

F

n

=

h

ct

2

1

h

c

t

1

= 1

t

1

S = h = 2 π ℏ

E = h /t

1

p = E /c = h /(ct

1

)

t

1

F

n

=

p

t

1

=

h

ct

2

1

F

n

=

6.62607015 × 10

−34

J·s

(299,792,458 m/s)(1 s)

2

= 2.21022 × 10

−42

N

A

i

= π r

2

i

G

m

i

= κ

i

π r

2

i

F

n

G

κ

i

of 24 73

Lorentz Invariance Check: Since contains and (invariants) and (invariant), is

Lorentz invariant. The cross-sectional area uses proper radius (invariant), and is

invariant. Therefore the entire expression for is Lorentz invariant, as required for rest mass.

3. Quantum Particle Physics: Master Equation

3.1 Master Equation Derivation

Starting from the mass formula and substituting :

Solving for yields the master equation:

This demonstrates the one-second interval embedded in matter's fundamental structure through

vacuum coherence. Since all quantities are Lorentz invariants, is Lorentz invariant.

3.2 Experimental Verification for Fundamental Particles

Proton: , = fine-structure constant:

Neutron: :

Electron: :

The remarkable consistency (0.99773--1.00500 seconds) provides compelling evidence for the

theory and vacuum origin of the Lorentz invariant one-second scale.

Note: The identical coupling for protons and neutrons reveals their deep connection

through strong and electromagnetic forces, while the electron's pure coupling suggests it

may represent the fundamental geometric unit of mass generation.

3.3 Physical Interpretation

Consider a particle's spacetime motion:

F

n

h

c

t

1

F

n

π r

2

i

r

i

G

m

i

F

n

m

i

= κ

i

π r

2

i

G

⋅

h

ct

2

1

t

1

t

1

=

r

i

m

i

⋅

πh

Gc

⋅ κ

i

t

1

κ

p

=

1

3α

2

α

t

1

=

0.833 × 10

−15

1.67262 × 10

−27

⋅

π ⋅ 6.62607 × 10

−34

(6.67430 × 10

−11

)(299,792,458)

⋅ 6256.33 = 1.00500 seconds

κ

n

=

1

3α

2

t

1

=

0.834 × 10

−15

1.675 × 10

−27

⋅

π ⋅ 6.62607 × 10

−34

(6.67430 × 10

−11

)(299,792,458)

⋅ 6256.33 = 1.00478 seconds

κ

e

= 1

t

1

=

2.81794 × 10

−15

9.10938 × 10

−31

⋅

π ⋅ 6.62607 × 10

−34

(6.67430 × 10

−11

)(299,792,458)

⋅ 1 = 0.99773 seconds

κ = 1/(3α

2

)

κ

e

= 1



V

spacetime

= (v

t

, v

s

) with ∥



V

spacetime

∥ = c

of 25 73

where is temporal velocity and spatial velocity. Accelerating a particle spatially rotates its

spacetime velocity vector, diverting motion from temporal to spatial dimensions. The normal

force resists this rotation, appearing as inertial resistance. This explains why mass is

proportional to energy: increasing spatial kinetic energy requires decreasing temporal "kinetic

energy," with resistance to this exchange manifesting as inertia, fundamentally mediated by

vacuum coherence.

4. Solar System Quantum Analog: Complete 1-Second

Invariance

Quantum-Cosmic Bridge: The same Lorentz invariant 1-second time scale governing

fundamental particles appears identically in solar system dynamics, creating a mathematical

bridge between quantum and cosmic scales.

4.1 Solar System Planck-Type Constant

We define a solar-system-scale analog to Planck's constant based on Earth's orbital kinetic energy

and the 1-second invariant:

where J, yielding:

4.2 Lunar Ground State and Exact 1-Second Invariance

The Moon's orbit exhibits quantum-like ground state behavior with exact 1-second characteristic

time:

Verification:

Lorentz Invariance Note: While is system-specific, the equation second

expresses a relationship between invariants: (action scale), (gravitational constant),

(rest mass), and (light speed). The resulting 1-second is thus Lorentz invariant.

4.3 Planetary Orbits as Quantum States

Planetary energy levels follow quantum-like formulas analogous to atomic orbitals:

where represents Earth's orbital quantum number and serves as a

normalized "charge" parameter (solar radius in lunar radius units).

v

t

v

s

F

n

ℏ

⊙

= (1 second) ⋅ K E

Earth

K E

Earth

=

1

2

M

e

v

2

e

≈ 2.7396 × 10

33

ℏ

⊙

≈ 2.7396 × 10

33

J·s

ℏ

2

⊙

GM

3

m

⋅

1

c

= 1 second

(2.7396 × 10

33

)

2

(6.67430 × 10

−11

) ⋅ (7.342 × 10

22

)

3

⋅

1

299,792,458

≈ 1.000 seconds

ℏ

⊙

ℏ

2

⊙

GM

3

m

⋅

1

c

= 1

ℏ

⊙

G

M

m

c

K E

e

= n ⋅

R

⊙

R

m

⋅

G

2

M

2

e

M

3

m

2ℏ

2

⊙

n = 3

R

⊙

/R

m

≈ 400

of 26 73

Verification for Earth (n=3): Predicted J matches actual orbital kinetic

energy with 99.5% accuracy.

4.4 Delocalization Time Computation

The Gaussian wavefunction's delocalization time provides further verification. For a wavepacket

initially localized with width , the spreading time is:

Taking and (lunar orbital diameter):

Half a year is:

The ratio or 97.2% agreement, showing the delocalization time

closely matches half the orbital period.

5. Mathematical Connection: Quantum and Cosmic Master

Equations

The Great Unification: The same mathematical form governs both quantum particles and

celestial mechanics, connected through the Lorentz invariant 1-second time scale.

5.1 Quantum Scale Master Equation

5.2 Solar System Scale Master Equation

5.3 Identical Mathematical Structure

Both equations share the identical form:

This demonstrates the same fundamental principle—a Lorentz invariant 1-second time scale—

governs both quantum particles and celestial bodies.

K E

e

≈ 2.739 × 10

33

d

τ =

m d

2

ℏ

m = m

moon

d = 2r

moon

τ =

m

moon

(2r

moon

)

2

ℏ

⊙

= 4

(7.34767 × 10

22

kg)(3.844 × 10

8

m)

2

2.8314 × 10

33

J·s

= 1.5338227 × 10

7

seconds

(1/2)(365.25)(24)(60)(60) = 1.57788 × 10

7

seconds

1.5338227/1.57788 = 0.972

t

(quantum)

1

=

r

p

m

p

⋅

πh

Gc

⋅

1

3α

2

= 1.00500 seconds

t

(solar)

1

=

R

m

M

m

⋅

π ℏ

⊙

Gc

⋅ κ

moon

= 1.000 seconds

t

1

=

characteristic length

characteristic mass

⋅

π × action constant

Gc

⋅ κ

of 27 73

5.4 Energy Quantization Comparison

Atomic scale (hydrogen atom):

Solar system scale (Earth-Moon):

Both exhibit characteristic quantum numbers and energy level quantization.

6. Fibonacci Optimization Across Scales

Different Fibonacci ratios optimize physical relationships at different scales, revealing

mathematical harmony across quantum and cosmic domains.

6.1 Quantum Scale Optimization (5/8 Ratio)

The proton radius relationship optimized by the Fibonacci ratio 5/8:

This yields near-perfect 1-second characteristic time:

The golden ratio approximation shows:

The Fibonacci sequence 0, 1, 1, 2, 3, 5, 8, 13,... converges to , with 5/8 = 0.625 providing the

closest two-term approximation.

6.2 Solar System Scale Optimization (2/3 Ratio)

The solar system Planck constant uses the 2/3 Fibonacci ratio:

E

n

= −

m

e

4

8ϵ

2

0

h

2

n

2

K E

n

= n ⋅

R

⊙

R

m

⋅

G

2

M

2

e

M

3

m

2ℏ

2

⊙

r

p

=

5

8

h

cm

p

r

p

=

5

8

6.62607 × 10

−34

(299,792,458)(1.67262 × 10

−27

)

= 0.8258821 × 10

−15

m

5

8

⋅

π r

p

α

4

Gm

3

p

1

3

⋅

h

c

= 1.0007 seconds

ϕ

r

p

= ϕ

h

cm

p

ϕ =

r

p

m

p

c

h

=

(0.833 × 10

−15

)(1.67262 × 10

−27

)(299,792,458)

6.62607 × 10

−34

= 0.6303866

ϕ

ℏ

⊙

= (hC )K E

e

hC = 1 second where C =

1

3

⋅

1

α

2

c

2

3

⋅

π r

p

Gm

3

p

ℏ

⊙

= (1.03351 s)(2.7396 × 10

33

J) = 2.8314 × 10

33

J·s

of 28 73

6.3 Earth-Moon Dynamics and the 24-Hour Day

The 24-hour Earth day emerges from lunar-terrestrial energy ratios:

Where EarthDay = 86,400 seconds and is Earth's axial tilt.

Note: The equation second suggests the characteristic rotation period

of Earth (24 hours) gives the characteristic time of 1 second when considering Moon's and

Earth's kinetic energies and Earth's axial tilt. The 24-hour day here is coordinate time, while the

1-second is Lorentz invariant.

7. Biological and Cosmological Connections

7.1 Carbon-Second Symmetry in Biochemistry

The 1-second invariant extends to biological chemistry through carbon-hydrogen relationships:

This 6:1 ratio establishes carbon as the temporal "unit cell" of biological chemistry, with its 6

protons exhibiting characteristic time of 1 second, while hydrogen (1 proton) shows 6-second

symmetry.

7.2 Cosmological Proton Freeze-Out

The 1-second scale was cosmologically imprinted during Big Bang nucleosynthesis:

This epoch corresponds to neutrino decoupling and proton-neutron ratio determination,

establishing fundamental particle properties.

7.3 Universal Lorentz Invariant Time Scale

The Complete Unification: The same Lorentz invariant time scale of approximately 1 second

appears in:

•

Quantum scale: Proton, neutron, electron characteristic times

•

Solar system scale: Lunar orbital ground state:

•

Biological scale: Carbon-hydrogen temporal symmetry

•

Cosmological scale: Big Bang nucleosynthesis timing

•

Human scale: 24-hour day emergence from celestial dynamics

K E

m

K E

e

(24 hours)cos(θ ) = 1.0 seconds

θ = 23.5

∘

KE

m

KE

e

(EarthDay)cos(θ ) = 1

1

6 protons

⋅

1

α

2

⋅

r

p

m

p

4πh

Gc

= 1 second (Carbon)

1

1 proton

⋅

1

α

2

⋅

r

p

m

p

4πh

Gc

= 6 seconds (Hydrogen)

t ≈

M

Pl

T

2

≈ 1.3 seconds at 1 MeV

ℏ

2

⊙

GM

3

m

⋅

1

c

= 1secon d

of 29 73

8. Discussion and Implications

8.1 Unification of Quantum Mechanics and Gravity

The theory represents a significant step toward unifying quantum mechanics and general

relativity. By identifying a quantum-gravitational interaction that generates inertial mass from

vacuum coherence, it bridges the conceptual gap between quantum theory's probabilistic nature

and gravity's geometric nature.

8.2 The Nature of Time

The emergence of the Lorentz invariant one-second scale from vacuum fluctuations suggests

time may be more fundamental than currently understood. Rather than being emergent, time

appears to have quantum structure with characteristic one-second scale, arising from vacuum

coherence properties.

8.3 Relativistic Consistency and Frame Independence

The recognition that is Lorentz invariant (not proper time) resolves potential relativistic

paradoxes. A proton moving at 0.99c has its proper time dilated but its characteristic

second remains unchanged—just as its rest mass and charge remain unchanged. This frame

independence is essential for the theory's consistency with special relativity.

8.4 Exoplanetary Predictions

The theory suggests habitable planets in other systems might exhibit similar 1-second

characteristics. Using the Vedic observation that Sun's diameter is ~108 Earth diameters and

Earth-Sun distance is ~108 solar diameters:

where AU from inverse square law. Applying to F through K spectral types

yields planet radii consistently near Earth's radius, suggesting optimally habitable planets may

have Earth-like size and mass due to life chemistry, atmospheric composition, and gravity

optimization.

8.5 Experimental Predictions

Fine-structure constant dependence: Any variation would manifest as changes in nucleon-

electron mass ratios.

Quantum gravity tests: The extremely weak normal force N suggests

possible experimental tests through ultra-sensitive force measurements or cosmological

observations.

Proton radius puzzle: The slight deviation from exactly 1 second in proton calculation (1.00500

s) may relate to the proton radius puzzle [5].

Vacuum coherence detection: The one-second coherence time should manifest as low-

frequency noise spectrum in ultra-precise fundamental constant measurements.

t

1

t

1

= 1

R

planet

= 2

R

2

⋆

r

planet

r

planet

= L

⋆

/L

⊙

α

F

n

≈ 2.21 × 10

−42

of 30 73

9. Conclusion

9.1 Summary of Key Results

Lorentz Invariant Time Scale:

Quantum Vacuum Fluctuations and Normal Force:

Mass Generation Mechanism:

Master Equation for All Scales:

Solar System Quantum Analog:

Planetary Quantum States:

Fibonacci-Optimized Predictions:

9.2 The Nature of Unification

This complete framework demonstrates that:

•

A Lorentz invariant one-second time scale governs phenomena across all physical

scales

•

Identical mathematical forms connect quantum particles and celestial mechanics

•

Fibonacci ratios optimize physical relationships at different scales (5/8 quantum, 2/3

cosmic)

•

The solar system exhibits quantum-like behavior with exact 1-second ground state

•

Biological complexity resonates with fundamental temporal patterns

t

1

= 1 second (Lorentz invariant)

F

n

=

h

ct

2

1

m

i

= κ

i

π r

2

i

F

n

G

t

1

=

r

i

m

i

⋅

πh

Gc

⋅ κ

i

ℏ

⊙

= (1 second) ⋅ K E

Earth

ℏ

2

⊙

GM

3

m

⋅

1

c

= 1 second

K E

e

= n ⋅

R

⊙

R

m

⋅

G

2

M

2

e

M

3

m

2ℏ

2

⊙

r

p

=

5

8

h

cm

p

= 0.8259 × 10

−15

m

of 31 73

•

The theory is fully relativistic with as Lorentz invariant, not frame-dependent proper

time

The appearance of the same Lorentz invariant time scale across all scales—from quantum

particles to planetary systems to biological organization—suggests we've identified a

fundamental natural principle. The One-Second Universe represents a cosmos structured around

a temporal invariant connecting quantum, cosmic, and biological realms through mathematical

harmony and empirical precision, while maintaining full consistency with special and general

relativity.

10. Defending the Theory

The theory finds:

works with proton, neutron, and electron radii. The natural constant is 1 second, much like

Newton's Universal Law where we don't explain why has its value—we measure it

and find it works, making it a Natural Law.

We derive this from the hypothesized normal force , giving:

, , ,

Derived from:

Equating left sides gives proton radius:

From and , proton frequency is . Setting yields

, thus .

Combining:

t

1

1 second =

r

i

m

i

⋅

πh

Gc

κ

i

F = G

Mm

r

2

G

F

n

= h /(ct

2

1

)

m

p

=

1

3α

2

⋅

π r

2

p

F

n

G

, m

e

=

π r

2

eClassical

F

n

G

, m

n

=

1

3α

2

⋅

π r

2

n

F

n

G

κ

p

= 1/(3α

2

)

κ

n

= 1/(3α

2

)

κ

e

= 1

r

e

= r

eClassical

(

1

6α

2

4πh

Gc

)

⋅

r

p

m

p

= 1 second

ϕ ⋅

π r

p

α

4

Gm

3

p

1

3

⋅

h

c

= 1 second

r

p

= ϕ ⋅

h

cm

p

E = h f

E = m

p

c

2

f

p

= m

p

c

2

/h

m

p

c

2

h

⋅

r

p

c

= ϕ

m

p

r

p

= ϕ ⋅

h

c

r

p

= ϕ ⋅

h

cm

p

of 32 73

Gives proton radius:

While we get very close to one second (1.0007 seconds) with the Fibonacci ratio 5/8, we also get

something very much in line with recent proton radius measurements ( ).

Relativistic Consistency: The equations use invariant quantities: rest masses , proper radii ,

and fundamental constants , , . Therefore the 1-second result is Lorentz invariant, ensuring

the theory's consistency with special relativity. A proton moving at relativistic speeds has the

same characteristic second, just as it has the same rest mass and charge.

References

[1] Beardsley, I. "The One-Second Universe: Quantum-Gravitational Unification Through a

Fundamental Temporal Invariant" (2025)

[2] Einstein, A. "On the Electrodynamics of Moving Bodies" Annalen der Physik 17, 891 (1905)

[3] Mach, E. "The Science of Mechanics" Open Court Publishing (1893)

[4] Ashby, N. "Relativity in the Global Positioning System" Living Reviews in Relativity 6, 1

(2003)

[5] Pohl, R. et al. "The size of the proton" Nature 466, 213–216 (2010)

[6] Xiong, W. et al. "A small proton charge radius from electron--proton scattering" Nature 575,

147–150 (2019)

[7] Bezginov, N. et al. "A measurement of the atomic hydrogen Lamb shift and the proton charge

radius" Science 365, 1007–1012 (2019)

[8] CODATA Internationally recommended values of the Fundamental Physical Constants (2018)

[9] Particle Data Group - Review of Particle Physics (2022)

[10] Planck Collaboration - Cosmological parameters (2018)

[11] Webb, J. K. et al. "Evidence for spatial variation of the fine structure constant" Physical

Review Letters 107, 191101 (2011)

[12] Misner, C. W., Thorne, K. S., & Wheeler, J. A. "Gravitation" Freeman (1973)

5

8

⋅

π r

p

α

4

Gm

3

p

1

3

⋅

h

c

= 1.0007 seconds

(

1

6α

2

4πh

Gc

)

⋅

r

p

m

p

= 1 second

r

p

=

5

8

h

cm

p

r

p

=

5

8

(6.62607 × 10

−34

)

(299,792,458)(1.67262 × 10

−27

)

= 0.8258821 × 10

−15

m

0.831 × 10

−15

m

i

r

i

h

G

c

t

1

= 1

of 33 73

[13] Rindler, W. "Relativity: Special, General, and Cosmological" Oxford University Press

(2006)

[14] Dirac, P. A. M. "The Principles of Quantum Mechanics" Oxford University Press (1930)

[15] Alexander Thom - "Megalithic Sites in Britain" (1967)

[16] Kepler Mission data on exoplanet characteristics

[17] ALMA observations of protoplanetary disks

[18] Big Bang Nucleosynthesis theoretical frameworks

[19] Biological timing and metabolic rate studies

[20] Fibonacci sequences in physical and biological systems

[21] Quantum gravity theoretical approaches

of 34 73

The Solar Solution Our solution of the wave equation for the planets gives the kinetic energy of the

Earth from the mass of the Moon orbiting the Earth, but you could formulate based on the Earth orbiting

the Sun. In our lunar formulation we had:

1.

We suggest the Moon perfectly eclipsing the Sun is a condition for the optimization of life on a planet.

This happens with our Solar System and we know the Moon holds the Earth at its inclination to its orbit

preventing extreme cold and extreme heat, allowing for the Seasons.

2.

Thus equation 1 becomes

3.

The kinetic energy of the Earth is

4.

Putting this in equation 3 gives the mass of the Sun:

5.

We recognize that the orbital velocity of the Moon is

6.

So equation 5 becomes

7.

This gives the mass of the Moon is

8.

Putting this in equation 1 yields

K E

e

= 3

R

⊙

R

m

⋅

G

2

M

2

e

M

3

m

2ℏ

2

⊙

R

⊙

R

m

=

r

e

r

m

K E

e

= 3

r

e

r

m

⋅

G

2

M

2

e

M

3

m

2ℏ

2

⊙

K E

e

=

1

2

⋅

GM

⊙

M

e

r

e

M

⊙

= 3r

2

e

⋅

GM

e

r

m

⋅

M

3

m

ℏ

2

⊙

v

2

m

=

GM

e

r

m

M

⊙

= 3r

2

e

⋅ v

2

m

⋅

M

3

m

ℏ

2

⊙

M

3

m

=

M

⊙

ℏ

2

⊙

3r

2

e

v

2

m

of 35 73

9.

We now multiply through by and we have

10.

The Planck constant for the Sun, , we will call , the subscript for Planck. We have

We write for the solution of the Earth/Sun system:

11.

We can write 11 as

12.

Where we say

Let us see how accurate our equation is:

=

K E

e

=

R

⊙

R

m

⋅

G

2

M

2

e

M

⊙

2r

2

e

v

2

m

M

2

e

/M

2

e

K E

e

=

R

⊙

R

m

⋅

G

2

M

4

e

M

⊙

2r

2

e

v

2

m

M

2

e

ℏ

⊙

L

p

L

p

= r

e

v

m

M

e

= r

e

v

m

M

e

= (1.496E11m)(1022m /s)(5.972E 24kg) = 9.13E 38kg ⋅

m

2

s

L

2

p

= r

2

e

v

2

m

M

2

e

= 7.4483E 77J ⋅ m

2

⋅ k g = 8.3367E 77kg

2

⋅

m

4

s

2

K E

e

=

R

⊙

R

m

⋅

G

2

M

4

e

M

⊙

2L

2

p

K E

e

=

R

⊙

R

m

⋅

G

2

M

4

e

M

⊙

2ℏ

2

⊙

ℏ

⊙

= 9.13E 38J ⋅ s

h

⊙

= 2 π ℏ

⊙

= 5.7365E 39J ⋅ s

K E

e

=

R

⊙

R

m

⋅

G

2

M

4

e

M

⊙

2L

2

p

R

⊙

R

m

(6.67408E − 11)

2

(5.972E 24kg)

4

(1.9891E 30kg)

2(8.3367E 77kg

2

⋅

m

4

s

2

)

R

⊙

R

m

(6.759E 30J )

R

⊙

R

m

=

6.957E 8m

1737400m

= 400.426

of 36 73

We have that the kinetic energy of the Earth is

Our equation has an accuracy of

Which is very good.

Let us equate the lunar and solar formulations:

This gives:

13.

We remember that

And since,

14.

K E

e

= 2.70655E 33J

K E

earth

=

1

2

(5.972E 24kg)(30,290m /s)

2

= 2.7396E 33J

2.70655E 33J

2.7396E 33J

= 98.79 %

K E

e

= n

R

⊙

R

m

⋅

G

2

M

2

e

M

3

m

2ℏ

2

⊙

K E

e

=

R

⊙

R

m

⋅

G

2

M

4

e

M

⊙

2ℏ

2

⊙

3

R

⊙

R

m

⋅

G

2

M

2

e

M

3

m

2ℏ

2

⊙

=

R

⊙

R

m

⋅

G

2

M

4

e

M

⊙

2L

2

p

L

p

=

M

2

e

M

⊙

M

3

m

3

⋅ ℏ

⊙

ℏ

⊙

= (hC )K E

e

hC = 1secon d

K E

e

=

1

2

M

e

v

2

e

2v

m

=

v

2

e

r

e

(1secon d )

M

2

e

M

⊙

M

3

m

3

of 37 73

Equation 14 becomes

15.

The condition of a perfect eclipse gives us another expression for the base unit of a second. is another

version of the Planck Constant, which is intrinsic to the the solar formulation as opposed to the lunar

formulation. We want to see what the ground state looks like and what its characteristic time is, if it is 1

second like it is for the lunar formulation. Looking at the equation for energy:

We see the ground state should be:

16.

And, it is equal to 1 second. You will notice where in the derivation for the energy we lost , we

have to put it in the ground state equation. The computation is:

M

2

e

M

⊙

M

3

m

3

=

(5.972E 24kg)

2

(1.9891E 30kg)

(7.34763E 22k g)

3

(1.732)

= 321,331.459 ≈ 321,331

1secon d = 2r

e

v

m

v

2

e

M

3

m

3

M

2

e

M

⊙

L

p

K E

e

=

R

⊙

R

m

⋅

G

2

M

4

e

M

⊙

2L

2

p

L

2

p

GM

2

e

M

⊙

⋅

3

c

= 1secon d

n = 3

(9.13E 38J ⋅ s)

2

(6.674E − 11)(5.972E 24kg)

2

(1.989E 30kg)

⋅

3

c

= 1.0172secon d s

of 38 73

Jupiter and Saturn We want to ﬁnd what the wave equation solutions are for Jupiter and Saturn because

they signiﬁcantly carry the majority of the mass of the solar system and thus should embody most clearly

the dynamics of the wave solution to the Solar System. We also show here how well the solution for the

Earth works, which is 99.5% accuracy.

I ﬁnd that as we cross the asteroid belt leaving behind the terrestrial planets, which are solid, and go to the

gas giants and ice giants, the atomic number is no longer squared and the square root of the the orbital

number moves from the numerator to the denominator. I believe this is because the solar system here

should be modeled in two parts, just as it is in theories of solar system formation because there is a force

other than just gravity of the Sun at work, which is the radiation pressure of the Sun, which is what

separates it into two parts, the terrestrial planets on this side of the asteroid belt and the gas giants on the

other side of the asteroid belt. The effect the radiation pressure has is to blow the lighter elements out

beyond the asteroid belt when the solar system forms, which are gases such as hydrogen and helium,

while the heavier elements are too heavy to be blown out from the inside of the asteroid belt, allowing for

the formation of the terrestrial planets Venus, Earth, and Mars. The result is that our equation has the

atomic number of the heavier metals such as calcium for the Earth, while the equation for the gas giants

has the atomic numbers of the gasses. We write for these planets

So, for Jupiter we have (And again using the maximum orbital velocity which is at perihelion):

Jupiter is mostly composed of hydrogen gas, and secondly helium gas, so it is appropriate that .

Our equation for Jupiter is

Where is the atomic number of hydrogen which is 1 proton, and for the orbital number of

Jupiter, . Now we move on to Saturn…

E =

Z

n

G

2

M

2

m

3

2ℏ

2

⊙

K E

j

=

1

2

(1.89813E 27kg)(13720m /s)

2

= 1.7865E 35J

E =

Z

H

5

(6.67408E − 11)

2

(1.89813E 27kg)

2

(7.347673E 22k g)

3

2(2.8314E 33)

2

E =

Z

H

5

(3.971E 35J ) = Z

H

(1.776E 35J )

Z

H

=

1.7865E 35J

1.776E 35J

= 1.006pr oton s ≈ 1.0pr oton s = h ydr ogen(H )

Z = Z

H

E

5

=

Z

H

5

G

2

M

2

j

M

3

m

2ℏ

2

⊙

Z

H

n = 5

K E

S

=

1

2

(5.683E 26kg)(10140m /s)

2

= 2.92E 34J

of 39 73

=

The equation for Saturn is then

It is nice that that Saturn would use Helium in the equation because Saturn is the next planet after Jupiter

and Jupiter uses hydrogen, and helium is the next element after hydrogen. As well, just like Jupiter, Saturn

is primarily composed of hydrogen and helium gas.

The accuracy for Earth orbit is

=

=2.727E33J

The kinetic energy of the Earth is

Which is very good, about 100% accuracy for all practical purposes. The elemental expression of the

solution for the Earth would be

Where

E =

Z

6

(6.67408E − 11)

2

(5.683E 26kg)

2

(7.347673E 22)

3

2(2.8314E 33)

2

Z

2.45

(3.5588E 34J ) = Z(1.45259E 34J )

Z(1.45259E 34J ) = (2.92E 34J )

Z = 2pr oton s = Heliu m(He)

E

6

=

Z

He

6

G

2

M

2

s

M

3

m

2ℏ

2

⊙

E

n

= n

R

⊙

R

m

G

2

M

2

e

M

3

m

2ℏ

2

⊙

R

⊙

R

m

=

6.96E 8m

1737400m

= 400.5986

E

3

= (1.732)(400.5986)

(6.67408E − 11)

2

(5.972E 24kg)

2

(7.347673E 22k g)

3

2(2.8314E 33)

2

K E

e

=

1

2

(5.972E 24kg)(30,290m /s)

2

= 2.7396E 33J

2.727E 33J

2.7396E 33J

100 = 99.5 %

E

3

= 3

Z

2

Ca

G

2

M

2

e

M

3

m

2ℏ

2

⊙

of 40 73

In this case the element associated with the Earth is calcium which is Z=20 protons.

R

⊙

R

m

→ Z

2

of 41 73

Modeling The Star System KOI-4878 We want to apply our wave theory for the planets to a star system

other than that of the Earth-Sun system, but that are similar, and fortunately we have discovered one such

a candidate star system. The G4V spectral type star KOI-4878 is in the constellation Draco with location

coordinates

RA: 19h 04m 54.7s

Dec: +50deg 00min 48.70sec

While M class stars are the most abundant in the galaxy and have longer life spans than the Sun, their

planets are thought to be tidally locked, their day is equal to their year, leaving them perpetually night on

one side of the planet and perpetually day on the other, only being cool enough for life in the twilight

region between night and day. K class stars show a lot of promise to host life on planets in their habitable

zones because they are far enough away from their star that they might not always become tidally locked,

while being more stable than the Sun and longer lived. It is easiest to detect planets and get data for these

M2V and K2V stars, but when you get to our Sun a G2V star, they are so bright they wash out the light of

their planets in the habitable zone a great deal. However, we have gotten data for a planet in the habitable

zone of a G4V star, about the same size, radius, mass, and luminosity as our Sun. And the planet in its

habitable zone has about the same radius, and perhaps mass as our Earth. All we need to do is to detect a

moon around its planet, and we will have veriﬁed my theory for habitable GV stars. We have yet

developed the technology to detect moons around a planet, but we are getting close to it, and we are going

to try to with the James Webb Space Telescope. This star is called KOI-4878 and its planet is

KOI-4878.01. Here is the information on it…

I have written a program in C that models star systems with our theory (Appendix 3). First we will do the

computation by hand so as to see how the program works. For KOI-4878 we will use some of the lowest

possible values, and ﬁnd we can get in the ball park. We will ﬁnd after running the program for higher

values of the data within the errors of the measurements for this star system, we can get close to this star

system. It becomes clear that with variations of parameters, this theory accounts for this star system if it

has a moon. And the moon can be similar to that of the Earth.

This star is the only candidate we have for an Earth-like planet in the habitable zone. We have detected

many around M-type red dwarfs because it is easy to detect planets around such abundant (the most

of 42 73

abundant) stars that are so faint that a transit lowers the light from the star by a high percentage. I say

candidate because there are rigid standards to consider them conﬁrmed. To be conﬁrmed you need to

detect them by methods other than transit (as this one was) like by measuring radial velocity (Changes in

velocity of the star due to being pulled on by the orbiting planet, by detecting red and blue shifts in the

star). It is hard to apply our theory for habitable planets to M-type stars because their habitable zones are

so close in that tidal forces from the star tidally lock the planet, so their rotation period gets slowed down

to their orbital period, leaving a gap in the data. Tidal forces weaken very rapidly with distance leaving

the Earth very unaffected by them. The tidal force gradient is proportional to , and tidal heating/

dissipation is proportional to . So at Earth, the effects are very small.

In order to apply the theory to other star systems, we have to be able to predict the radius of the habitable

planet, presumably in the n=3 orbit. I found the answer to be in the Vedic literature of India. They noticed

that the diameter of the Sun is about 108 times the diameter of the Earth and that the average distance

from the Sun to the Earth is about 108 solar diameters, with 108 being a signiﬁcant number in Yoga. So I

wrote the equivalent:

radius of the star. The surprising result I found was, after applying it to the stars of all spectral types

from F through K, with their different radii and luminosities (the luminosities determine , the

distances to the habitable zones), that the radius of the planet always came out about the same, about the

radius of the Earth. This may suggest optimally habitable planets are not just a function of their distance

from the star, which determines their temperature, but are functions of their size and mass probably

because they are good for life chemistry atmospheric composition, and gravity when they are the size and

mass of the Earth.

In order to get , the distance of the habitable planet from the star, we use the inverse square law for

luminosity of the star. If the Earth is in the habitable zone, and if the star is one hundred times brighter

than the Sun, then by the inverse square law the distance to the habitable zone of the planet is 10 times

that of what the Earth is from the Sun. Thus we have in astronomical units the habitable zone of a star is

given by:

the luminosity of the star, and the luminosity of the Sun.

1/r

3

1/r

6

R

planet

= 2

R

2

⋆

r

planet

R

⋆

r

planet

r

planet

r

planet

=

L

⋆

L

⊙

AU

L

⋆

L

⊙

of 43 73

A G4V star on average has a mass of 0.985, a radius of 0.991, a luminosity of 0.91 (Sun=1). Since the

above data has a large margin of error taking it to a range of 0.88-1.138 solar masses (avg. 0.9325) we

will use the average for the G4V spectral type that it is, which is 0.985 solar masses. And since the radius

is in the range is 1.072-1.19 solar radii, (avg. 1.131) we will use the average again for its spectral class

G4V which is 0.991. This gives

The mass of the star being taken to be 0.985 solar masses, we have, if the orbit of the planet is close to

circular:

=

This is to see if the period predicted is close to the period measured, which it is because the measured

value is 449.015 days. That is 97.5%. This is good because we used a circular orbit approximation and

average values for G4V stars. Let us compute the kinetic energy of this planet:

Compared to that of Earth, which is 29,784m/s. The mass of the planet we will take to be 0.92 that of

Earth as recommended by Wikipedia because it has a range of 0.66-1.18 Earth masses. That is

[(0.92)(5.972E24kg)]=5.49424E24kg

The kinetic energy is, then:

We now compute , the Planck-type constant for this star system. We use

Where is the exponent in

R

planet

= 2

R

2

⋆

r

planet

= 2

[(0.991)(6.96E 8m)]

2

(1.496E11m /AU )(1.125)

= 5.6534E 6m =

(6.378E6m)

(5.6534E6m)

= 1.1282Ear th Ra dii

r

planet

=

L

⋆

L

⊙

AU =

0.91

1

AU = 0.9539392AU = 1.4271E11m

T

2

=

4π

2

GM

a

3

=

(39.4784)

(6.674E − 11)(0.991)(1.989E 30kg)

[(1.125)(1.496E11m)]

3

(3.001E − 19)(1.683E11m)

3

= 1.430600E15

T = 3.7823E 7secon d s = 437.77d a ys

v =

GM

r

=

(6.674E − 11)(1.971E 30kg)

(1.683E11m)

= 27,957.244m /s

K E =

1

2

Mv

2

=

1

2

(5.4942E 24kg)(27,957.244)

2

= 2.147E 33J

ℏ

⋆

L

earth

ℏ

⊙

= p

p

of 44 73

The pressure gradient for the protoplanetary disc from which the planets formed (See Appendix 1). We

have

Since Mars is further out and has a day of close to Earth’s 24 hours, and since Venus doesn’t have this

because it is closer to the Sun and greatly slowed down by tidal forces, we will guess for an Earth-sized

planet like this one, its day is 24hrs=86,400sec because we are computing as if this planet hosts life, and a

fast rotation, keeps the planet cool, but it can’t be so fast that the nights and days are to short for life to

function (hunt, build, etc…):

=

For G4V stars the typical range of is p=1.6-2.0 for the exponent in the pressure gradient. We will choose

2.0 since it is closest to that of Earth, which is 2.5:

Compared to that of Earth, which is . Thus we have the characteristic time of this

planet is

Our theory says that

So the mass of the Moon of this planet is:

=6.4989E22kg~6.5E22kg

P(R) = P

0

(

R

0

)

−L

ear th

ℏ

⊙

L

planet

=

4

5

π M

p

f

p

R

2

p

L

p

=

4

5

π (5.4942E 24k g)

1

(86400secon d s)

(5.6534E6m))

2

5.108E 33J ⋅ s

(5.108E 33J ⋅ s)

ℏ

⋆

= 2.0

ℏ

⋆

= 2.554E 33J ⋅ s

ℏ

⊙

: 2.8314E 33J ⋅ s

t

c

=

ℏ

⋆

K E

p

=

(2.554E 33J ⋅ s)

(2.147E 33J )

= 1.1877secon d ≈ 1secon d

ℏ

2

⊙

GM

3

m

⋅

1

c

= 1secon d

M

3

m

=

(2.554E 33J ⋅ s)

2

(6.674E − 11)(299,729,458m /s)(1.1877secon d s)

of 45 73

Compared to that of the Earth’s moon, 7.347673E22kg. The orbital radius of the Earth’s moon seems to

be governed by the relative masses of the heavy metallic elements gold (Au) and silver (Ag). We will

guess this holds here, which is a similar type of a star system.

It is given by the ratio of silver (Ag) to gold (Au) by molar mass is equal to . The radius of the

planet’s moon we suggested is given by a perfect eclipse:

Compared to that of the Moon, which is 3.84E8m. From this we have the radius of the Moon:

Compared to that of the Moon, which is 1.7374E6m. Now to get the density of the Moon…

Compared to the Earth moon 3.34 g/cm3. The Earth’s moon is consists of silicates for the surface regolith,

which is porous, with a low density starting at 1.5g/cm3 to solid lunar rock and mantle of 3.17-3.22g/cm3

and 3.22-3.34g/cm3. This moon could exist with a smaller iron core and higher proportions of lighter

silicates. We want to compute the orbital kinetic energy of this moon.

Compared to that of the Earth’s moon, which is 1022m/s

Where that with the Earth’s moon it is 3.428E28J using its orbital velocity at aphelion, which is 966m/s.

We can now computer the PlanetDay characteristic time:

r

m

= R

⊙

Ag

Au

=

R

⋆

(1.8)

r

m

/R

⊙

R

⋆

R

m

=

r

p

r

m

r

m

= R

⋆

Ag

Au

= R

⋆

/(1.8) =

(0.991)(6.96E 8m)

1.8

= 3.832E 8m

R

m

= R

⋆

r

m

r

p

= (6.957E 8m)

3.832E 8m

1.496E11m

= 1.782E6m

V

m

=

4

3

π R

3

m

=

4

3

π (1.782E 6 m)

3

= 2.37E19m

3

ρ

m

=

6.5E 22k g

2.37E19m

3

= 2742.62 k g /m

3

≈ 2.74262g /c m

3

v =

GM

r

=

(6.674E − 11)(5.49424E 24kg)

(3.832E 8m)

= 978.21m /s

K E

m

=

1

2

(6.5E 22k g)(978.2m /s)

2

= 3.12E 28J

K E

m

K E

p

(Pl a n et D a y)cos(θ ) ≈ 1.0secon d s

of 46 73

=

This is close to the characteristic time for star system, which we found was

Running our program for the Earth to verify its accuracy, we have

What is the radius of the star in solar radii? 1

What is the mass of the star in solar masses? 1

What is the luminosity of the star in solar luminosities? 1

Do you want us to compute the planet radius, 1=yes, 0=no? 1

What is the mass of the planet in Earth masses? 1

What is the planet day in Earth days? 1

That is 86400.000000 seconds

What is p the pressure gradient exponent of the protoplanetary disc? 2.5

Angular Momentum of Planet: 7.187518 E33

PlanetYear: 0.999888 years

PlanetYear: 31554074.000000 seconds

planet orbital velocity: 29788.980469 m/s

planet mass: 5.972000 E24 kg

planet mass: 1.000000 Earth masses

planet radius 6432306.000000 meters

planet radius: 1.008515 Earth Radii

planet orbital radius: 1.496000 E11 m

planet orbital radius: 1.000000 Earth distances

planet KE: 2.649727 E33 J

planet density: 5.357135 g/cm3

hbarstar: 2.875007 E33 Js

characteristic time: 1.085020 seconds

Orbital Radius of Moon: 3.853556 E8 m

Orbital Radius of Moon: 1.003530 Moon Distances

Radius of Moon: 1.793707 E6 m

Radius of Moon: 1.032408 Moon Radii

Mass of Moon: 7.247882 E22 kg

Mass of Moon 0.986419 Moon Masses

density of moon: 2.998263 g/cm3

Orbital Velocity of Moon: 1017.002930 m/s

PlanetDay Characteristic Time: 1.120820 seconds

Lunar Orbital Period: 2380778.000000 seconds

Lunar Orbital Period: 27.555302 days

Program ended with exit code: 0

(3.12E 28J )

(2.147E 33J )

(86,400s)cos(23.5

∘

) = 1.15secon d s

t

c

=

ℏ

⋆

K E

p

=

(2.554E 33J ⋅ s)

(2.147E 33J )

= 1.1877secon d ≈ 1secon d

of 47 73

We see it works great. Characteristic time is 1.085 seconds, Lunar mass is

0.98 moons, its density is 2.998g/cm3 close that of the Earth’s moon. The

PlanetDay characteristic time is 1.12 seconds.

We run it for KOI-4878

What is the radius of the star in solar radii? 1

What is the mass of the star in solar masses? 1

What is the luminosity of the star in solar luminosities? 1.3

Do you want us to compute the planet radius, 1=yes, 0=no? 1

What is the mass of the planet in Earth masses? 1

What is the planet day in Earth days? 1

That is 86400.000000 seconds

What is p the pressure gradient exponent of the protoplanetary disc? 2.5

Angular Momentum of Planet: 5.528859 E33

PlanetYear: 1.217332 years

PlanetYear: 38416076.000000 seconds

planet orbital velocity: 27897.789062 m/s

planet mass: 5.972000 E24 kg

planet mass: 1.000000 Earth masses

planet radius 5641505.000000 meters

planet radius: 0.884526 Earth Radii

planet orbital radius: 1.705703 E11 m

planet orbital radius: 1.140175 Earth distances

planet KE: 2.323964 E33 J

planet density: 7.940497 g/cm3

hbarstar: 2.211544 E33 Js

characteristic time: 0.951626 seconds

Orbital Radius of Moon: 3.853556 E8 m

Orbital Radius of Moon: 1.003530 Moon Distances

Radius of Moon: 1.573184 E6 m

Radius of Moon: 0.905482 Moon Radii

Mass of Moon: 6.356811 E22 kg

Mass of Moon 0.865146 Moon Masses

density of moon: 3.897743 g/cm3

Orbital Velocity of Moon: 1017.002930 m/s

PlanetDay Characteristic Time: 1.120820 seconds

Lunar Orbital Period: 2380778.000000 seconds

Lunar Orbital Period: 27.555302 days

Program ended with exit code: 0

We find to get the orbital period the planet has, one solution is to run it

at mass and size of the Sun (which is within the errors for its actual value)

and use p=2.5 like it is for the Sun, but the luminosity 1.3 solar

luminosities.

This gives an orbital period (planet year) of 444.63 days

of 48 73

Running it again for KOI-4878 varying parameters…

What is the radius of the star in solar radii? 1.072

What is the mass of the star in solar masses? 1

What is the luminosity of the star in solar luminosities? 1.3

Do you want us to compute the planet radius, 1=yes, 0=no? 1

What is the mass of the planet in Earth masses? 1

What is the planet day in Earth days? 1

That is 86400.000000 seconds

What is p the pressure gradient exponent of the protoplanetary disc? 2.5

Angular Momentum of Planet: 7.301542 E33

PlanetYear: 1.217332 years

PlanetYear: 38416076.000000 seconds

planet orbital velocity: 27897.789062 m/s

planet mass: 5.972000 E24 kg

planet mass: 1.000000 Earth masses

planet radius 6483127.000000 meters

planet radius: 1.016483 Earth Radii

planet orbital radius: 1.705703 E11 m

planet orbital radius: 1.140175 Earth distances

planet KE: 2.323964 E33 J

planet density: 5.232137 g/cm3

hbarstar: 2.920617 E33 Js

characteristic time: 1.256739 seconds

Orbital Radius of Moon: 4.131012 E8 m

Orbital Radius of Moon: 1.075784 Moon Distances

Radius of Moon: 1.807878 E6 m

Radius of Moon: 1.040566 Moon Radii

Mass of Moon: 6.974273 E22 kg

Mass of Moon 0.949181 Moon Masses

density of moon: 2.817760 g/cm3

Orbital Velocity of Moon: 982.256287 m/s

PlanetDay Characteristic Time: 1.147099 seconds

Lunar Orbital Period: 2642476.250000 seconds

Lunar Orbital Period: 30.584215 days

Program ended with exit code: 0

This gives an orbital period (planet year) of 444 days

Importantly, since the Moon is pivotal to our theory, the important thing is

we get its density so its composition is right and close to its orbital

period of 449 days. We get it is in the range of

Range: 2.818g/cm3-3.898g/cm3. 444.63 days Characteristic time: 0.95s

Average: 3.356g/cm3. 444 days. Characteristic time: 1.25674s

of 49 73

The density of the Earth’s moon is 3.34g/cm3. We get about exactly this in the average. That of the Earth

is 5.52g/cm3, we got 5.23g/cm3 for this planet in the second running or the program, but 7.94g/cm3,

about half that of the planet Mercury (13.6g/cm3). Clearly, with variation of parameters, we can get this

star system. We want the moon to be right because it is believed it is very important to have a moon

orbiting the planet if the planet is to be high functioning in its habitability because it prevents hot and cold

weather extremes. It allows for stable conditions over long periods to give life a chance to evolve into

something sophisticated, like intelligent life. It does this by holding the planet at its inclination to it orbit,

which for the earth is about 1/4 of a right angle (23.5 deg) which is what we used here, the same what it is

for the Earth.

The constellation Draco, which is

Latin for “the Dragon” is a large

winding constellation visible all

year in the Northern Hemisphere.

Since it is near the North Star,

Polaris, it goes around it near

the Little Dipper and Big Dipper

always high in the sky. The

brightest star in it is alpha

Draconis, common name Thuban,

which was the pole star when the

Egyptian Pyramids were being

built, and were thus aligned with

it. It will be the pole star again

in 21000 AD due to the Earth’s

precession. It was the pole star

from 3942 BC to 1793 BC.

I have applied this program to a wide range of stars, using their average

values for stellar mass, stellar radius, and stellar luminosities. We see the

characteristic times of about 1 second intersect around spectral class GV

stars like our Sun. Here we show such results for F5V stars down to G3V stars

(which are near to the Sun) down to as low in mass, luminosity, and radius

such as K3V stars. We see using our equation

which is where in the program we give the option to compute the planet’s

radius, that it always returns something close to the Earth radius. We use

the equation for that. We see the characteristic time of 1 second for the

star system intersects with the PlanetDay characteristic time of 1 second

around G-type stars like the Sun, putting them inline with the proton,

electron, and neutron.

The results are…

R

planet

= 2

R

2

⋆

r

planet

of 50 73

F5V Star

What is the radius of the star in solar radii? 1.473

What is the mass of the star in solar masses? 1.33

What is the luminosity of the star in solar luminosities? 3.63

Do you want us to compute the planet radius, 1=yes, 0=no? 1

What is the mass of the planet in Earth masses? 1

What is the planet day in Earth days? 1

That is 86400.000000 seconds

What is p the pressure gradient exponent of the protoplanetary disc? 2.4

Angular Momentum of Planet: 9.321447 E33

PlanetYear: 2.280109 years

PlanetYear: 71954776.000000 seconds

planet orbital velocity: 24888.847656 m/s

planet mass: 5.972000 E24 kg

planet mass: 1.000000 Earth masses

planet radius 7325190.000000 meters

planet radius: 1.148509 Earth Radii

planet orbital radius: 2.850263 E11 m

planet orbital radius: 1.905256 Earth distances

planet KE: 1.849692 E33 J

planet density: 3.627237 g/cm3

hbarstar: 3.883936 E33 Js

characteristic time: 2.099775 seconds

Orbital Radius of Moon: 5.676287 E8 m

Orbital Radius of Moon: 1.478200 Moon Distances

Radius of Moon: 2.042695 E6 m

Radius of Moon: 1.175720 Moon Radii

Mass of Moon: 7.107576 E22 kg

Mass of Moon 0.967323 Moon Masses

density of moon: 1.990782 g/cm3

Orbital Velocity of Moon: 837.955261 m/s

PlanetDay Characteristic Time: 1.068920 seconds

Lunar Orbital Period: 4256210.000000 seconds

Lunar Orbital Period: 49.261688 days

Program ended with exit code: 0

of 51 73

G3V Star

What is the radius of the star in solar radii? 1.002

What is the mass of the star in solar masses? 0.99

What is the luminosity of the star in solar luminosities? 0.98

Do you want us to compute the planet radius, 1=yes, 0=no? 1

What is the mass of the planet in Earth masses? 1

What is the planet day in Earth days? 1

That is 86400.000000 seconds

What is p the pressure gradient exponent of the protoplanetary disc? 2.1

Angular Momentum of Planet: 7.393050 E33

PlanetYear: 0.989814 years

PlanetYear: 31236148.000000 seconds

planet orbital velocity: 29789.738281 m/s

planet mass: 5.972000 E24 kg

planet mass: 1.000000 Earth masses

planet radius 6523625.500000 meters

planet radius: 1.022833 Earth Radii

planet orbital radius: 1.480965 E11 m

planet orbital radius: 0.989950 Earth distances

planet KE: 2.649862 E33 J

planet density: 5.135297 g/cm3

hbarstar: 3.520500 E33 Js

characteristic time: 1.328560 seconds

Orbital Radius of Moon: 3.861263 E8 m

Orbital Radius of Moon: 1.005537 Moon Distances

Radius of Moon: 1.819172 E6 m

Radius of Moon: 1.047066 Moon Radii

Mass of Moon: 7.754257 E22 kg

Mass of Moon 1.055335 Moon Masses

density of moon: 3.074905 g/cm3

Orbital Velocity of Moon: 1015.987427 m/s

PlanetDay Characteristic Time: 1.196672 seconds

Lunar Orbital Period: 2387924.250000 seconds

Lunar Orbital Period: 27.638012 days

Program ended with exit code: 0

of 52 73

K3V Star

What is the radius of the star in solar radii? 0.755

What is the mass of the star in solar masses? 0.78

What is the luminosity of the star in solar luminosities? 0.28

Do you want us to compute the planet radius, 1=yes, 0=no? 1

What is the mass of the planet in Earth masses? 1

What is the planet day in Earth days? 1

That is 86400.000000 seconds

What is p the pressure gradient exponent of the protoplanetary disc? 1.5

Angular Momentum of Planet: 8.340819 E33

PlanetYear: 0.435785 years

PlanetYear: 13752343.000000 seconds

planet orbital velocity: 36167.078125 m/s

planet mass: 5.972000 E24 kg

planet mass: 1.000000 Earth masses

planet radius 6929175.500000 meters

planet radius: 1.086418 Earth Radii

planet orbital radius: 0.791609 E11 m

planet orbital radius: 0.529150 Earth distances

planet KE: 3.905860 E33 J

planet density: 4.285367 g/cm3

hbarstar: 5.560546 E33 Js

characteristic time: 1.423642 seconds

Orbital Radius of Moon: 2.909435 E8 m

Orbital Radius of Moon: 0.757665 Moon Distances

Radius of Moon: 1.932263 E6 m

Radius of Moon: 1.112158 Moon Radii

Mass of Moon: 10.277222 E22 kg

Mass of Moon 1.398704 Moon Masses

density of moon: 3.400867 g/cm3

Orbital Velocity of Moon: 1170.438843 m/s

PlanetDay Characteristic Time: 1.428032 seconds

Lunar Orbital Period: 1561850.125000 seconds

Lunar Orbital Period: 18.076969 days

Program ended with exit code: 0

of 53 73

PlanetDay characteristic time:

Characteristic time:

We name the spectral types with number for input according to the following scheme.

F5V is 1.5, F6V is 1.6, F7V is 1.7,…G0V is 2.0, G1V is 2.1,…

We see the tendency is towards characteristic time and planetary characteristic time intersecting at a

minimum in the area GV stars (G3V=2.3) like our Sun, where star systems come in line with the

characteristic time of the proton, electron, and neutron. This may be the place of optimal habitability. GV

stars come in line with the electron, proton, and neutron characteristic time given by:

K E

m

K E

e

(Pl a n etDa y)cos(23.5

∘

) = 1secon d

ℏ

2

⊙

GM

3

m

⋅

1

c

= 1secon d

1secon d =

r

i

m

i

⋅

πh

G c

κ

i

of 54 73

Appendix 1 Pressure Gradient of the Protoplanetary Disk

We would like to see how our wave solution for the solar system figures into the classical

analytic theory of the formation of our solar system.The protoplanetary disc that evolves into the

planets has two forces that balance its pressure, the centripetal force of the gas disc due to its

rotation around the protostar and the inward gravitational force on the disc from the

protostar , and these are related by the density of the gas that makes up the disc. The

pressure gradient of the disc in radial equilibrium balancing the inward gravity and outward

centripetal force is

1.

We can solve this for pressure in the protoplanetary disc as a function of r, distance from the

star, as follows: Assume the gas is isothermal, meaning the temperature T is constant so we can

relate pressure and density with

Where is the speed of sound in the gas which depends on its temperature. We take the gas to

be in nearly Keplerian rotation. That is the rotation is given by Newtonian gravity:

And we take into account that the rotational velocity is slowed down by gas pressure using the

the parameter which is less than one:

We can say for a protoplanetary disc like that from which our solar system originated that its

density varies with radius as a power law:

is the reference density at and s is the power law exponent. We can write

.

We have from 1:

2.

Since , we have that which gives from 2:

v

2

ϕ

/r

GM

⋆

/r

2

ρ

d P

dr

= − ρ

(

GM

⋆

r

2

−

v

2

ϕ

r

)

P = c

2

s

ρ

c

s

v

K

=

GM

⋆

r

η

v

ϕ

= v

K

(1 − η)

ρ(r) = ρ

0

(

r

0

)

−s

ρ

0

r

0

v

2

ϕ

= v

2

K

(1 − η)

2

≈

GM

⋆

r

(1 − 2η)

d P

dr

= − ρ

(

GM

⋆

r

2

⋅ 2η

)

P = c

2

s

ρ

d P/dr = c

2

s

dρ /dr

of 55 73

We integrate both sides:

And, we have

3.

We take

as small because is small and r is large so we can make the approximation . We

have

4.

What we can get out of this is since the deviation parameter, , is given by

5. and

6.

Where, is the Boltzmann constant, is the molecular weight of

hydrogen, and is the mass of hydrogen is basically the mass of a proton is 1.67E-27kg. Since

for a protoplanetary cloud at Earth orbit T is around 280 degrees Kelvin we have

dρ

ρ

= −

2ηGM

⋆

c

2

s

r

2

dr

∫

ρ

0

dρ

ρ

= −

2ηGM

⋆

c

2

s

r

2

∫

r

0

dr

ln

(

ρ

0

)

=

2ηGM

⋆

c

2

s

(

1

r

0

−

1

r

)

ρ(r) = ρ

0

ex p

2ηGM

⋆

c

2

s

(

1

r

0

−

1

r

)

P

0

= c

2

s

ρ

0

ex p

2ηGM

⋆

c

2

s

(

1

r

0

−

1

r

)

2ηGM

⋆

c

2

s

(

1

r

0

−

1

r

)

η

e

x

≈ 1 + x

P

r

≈ P

0

1 +

2ηGM

⋆

c

2

s

(

1

r

0

−

1

r

)

P

0

= c

2

s

ρ

0

η

η = −

1

2

(

c

s

v

K

)

2

dln P

dln R

c

s

=

k

B

T

μm

H

k

B

= 1.38E − 23J/K

μ ≈ 2.3

m

H

of 56 73

Typically in discs the pressure decreases with radius as a power law

Where , so

7.

So, essentially, by the chain rule

to clarify things. The reason 7 is significant is that equation

Where

c

s

= 1k m /s

P(R) ∝ R

−q

q ≈ 2.5

dln P

dln R

≈ − 2.5

η = −

1

2

(

1k m /s

30k m /s

)

2

(−2.5) = 1.5E − 3

dln P

dln R

=

dln P

d R

⋅

d R

dln R

=

1

P

d P

d R

⋅ R =

R

P

d P

d R

L

earth

ℏ

⊙

=

7.05E 33

2.8314E 33

= 2.4899 ≈ 2.5 = 2

1

2

L

earth

=

4

5

π M

e

f

e

R

2

e

ℏ

⊙

= 2.8314E 33J ⋅ s

λ

moon

c

=

ℏ

2

⊙

GM

3

m

⋅

1

c

= 1.0secon d s

of 57 73

Appendix 2: Deriving the Delocalization Time From a Gaussian Wave Packet

In order to show that our hypothesis is right, we solve the wave equation for a Gaussian wave

packet and determine the delocalization time, . If it is about six months, the time it takes the

Earth to delocalize (travel its orbital diameter), using the Moon playing the role of the mass of

an electron and our as above to describe the Earth, then the hypothesis can be taken as

correct, and we can solve the whole system for the Earth/Moon/Sun system from the rest of

the equations in the hydrogen atom solution to the Schrodinger wave equation, which is in

spherical coordinates:"

"

The delocalization time of a particle, molecule, or mass in general, , is the time it takes a

particle to delocalize. If we want to apply our wave equation theory of the Solar System to this

concept, then the delocalization time should be the time for the Earth to travel the diameter of

it’s orbit, which would be half a year (about six months). In order to derive the delocalization

time we must consider a Gaussian wave packet…"

"

The Gaussian wavefunction in position space is"

τ

ℏ

⊙

−

ℏ

2

2m

[

1

r

2

∂

∂r

(

r

2

∂

∂r

)

+

1

r

2

sinθ

∂

∂θ

(

sinθ

∂

∂θ

)

+

1

r

2

sin

2

θ

∂

2

∂ϕ

2

]

ψ + V(r)ψ = E ψ

τ

of 58 73

"

It’s Fourier wave decomposition is"

"

We use the Gaussian integral identity (integral of quadratic)"

"

We ﬁnd via the inverse Fourier transform. It is"

"

Substitue :"

"

This is of the form:"

"

, "

Using"

"

ψ (x,0) = Ae

−

x

2

2d

2

ψ (x,0) = Ae

−

x

2

2d

2

=

∫

dp

2π ℏ

ϕ( p)e

i

ℏ

px

∫

∞

−∞

e

−a x

2

+bx

d x =

π

a

e

b

2

4a

ϕ( p)

ϕ( p) =

∫

∞

−∞

d x ψ (x,0)e

−

i

ℏ

px

ψ (x,0)

ϕ( p) = ϕ( p) = A

∫

∞

−∞

e

−

x

2

2d

2

e

−

i

ℏ

[ px]

d x

∫

e

{−a x

2

−bx}

d x

a =

1

2d

2

b =

ip

ℏ

∫

∞

−∞

e

−a x

2

+bx

d x =

π

a

e

b

2

4a

ϕ( p) = A

π

1/(2d

2

)

ex p

−

p

2

ℏ

2

4 ⋅

1

2d

2

b

2

4a

= −

p

2

d

2

2ℏ

2

of 59 73

"

We have to ﬁnd the normalization constant, A, because the probability has to be 1 at its

maximum. We have"

"

We now consider the evolution of a free particle. For a free particle"

"

The time evolution in free space is"

"

Substitute:"

"

Factor out the term and we have"

"

The integral is then,"

ϕ( p) = Ad 2πexp

(

−

p

2

d

2

2ℏ

2

)

∫

∞

−∞

|

ψ (x,0)

|

2

= 1

|

ψ (x,0)

|

2

=

|

A

|

2

e

−x

2

/d

2

∫

∞

−∞

e

{−x

2/

d

2}d x

= d π

A = (πd

2

)

−1/4

H =

p

2

2m

ϕ( p, t) = ϕ( p)e

−i

p

2

2m

t/h

= ϕ( p)e

−i

p

2

t

2mℏ

ψ (x, t) =

∫

dp

2π ℏ

ϕ( p)e

{

i

ℏ

px}

e

{−i

p

2

t

2mℏ

}

ϕ( p) = Ad 2πexp

(

−

p

2

d

2

2ℏ

2

)

ψ (x, t) = A

d 2π

2π ℏ

∫

∞

−∞

dpexp

[

−

p

2

d

2

2ℏ

2

− i

p

2

t

2mℏ

+

ipx

ℏ

]

p

2

α =

d

2

2ℏ

2

+

it

2mℏ

=

1

2ℏ

2

[

d

2

+

itℏ

m

]

of 60 73

, Re(a)>0"

, "

"

, , , , "

"

We determine the probability density . We have"

"

, , "

"

Because we multiplied the top and bottom of by and took its real part. We have"

"

∫

∞

−∞

e

{−αp

2

+bp}

dp =

π

α

e

b

2

/(4a)

a = α

b = i x /ℏ

A

d 2π

2π ℏ

π

α

=

Ad

ℏ

1

2α

ψ (x, t) =

Ad

ℏ 2α

ex p

[

−

x

2

/ℏ

2

4α

]

τ =

m d

2

ℏ

m d

2

=

1

τ

ℏ/m

d

2

=

1

τ

itℏ/m

d

2

=

it

τ

ψ (x, t) =

Ad

ℏ 2α

ex p

[

−

x

2

2d

2

(1 + it /τ)

]

|

ψ (x, t)

|

2

|

ψ (x, t)

|

2

=

|

Ad

ℏ 2α

|

2

ex p

[

−

x

2

2d

2

⋅ 2Re

(

1

1 + it /τ

)

]

|

ex p(−Bx

2

)

|

2

= exp(−2ReBx

2

)

B =

1

2d

2

(1 + it /τ)

ReB =

1

2d

2

⋅

1

1 + t

2

/τ

2

2ReB = 2 ⋅

1

2d

2

(1 + t

2

/τ

2

)

=

1

d

2

(1 + t

2

/τ

2

)

B

1 − it /τ

1

1 + i

t

τ

⋅

1

1 − i

t

τ

=

1

1 +

t

2

τ

2

|

ψ (x, t)

|

2

∝ exp

[

−

x

2

d

2

(1 + t

2

/τ

2

)

]

|

ψ (x, t)

|

2

=

[

−

x

2

d

2

⋅

1

(1 + t

2

/τ

2

)

]

τ =

m d

2

ℏ

of 61 73

is the delocalization distance, which for instance could be the width of an atom. is the

delocalization time, the average time for say an electron to traverse the diameter of the atom and

even leave it, to delocalize. If we substitute for our , and say that the delocalization distance

uses for the Moon, the width of the Earth orbit, we should get a half a year for the delocalization

time, the time for the Moon and Earth to traverse the diameter of their orbit around the Sun. We

have

Where is the mass of the Moon, and is the orbital radius of the Moon. We have

Now let’s compute a half a year…

(1/2)(365.25)(24)(60)(60)=15778800 seconds

So we see our delocalization time is very close to the half year over which the Earth and Moon

travel from one position to the opposite side of the Sun. The closeness is

d

τ

ℏ

⊙

τ =

m

moon

(2r

moon

)

2

ℏ

⊙

m

moon

r

moon

τ = 4

(7.34767E 22kg)(3.844E8m)

2

2.8314E 33J ⋅ s

= 15338227secon d s

15338227

15778800

100 = 97.2 %

of 62 73

Appendix 3: The Program For Modeling Star Systems

//

// main.c

// modelsystem

//

// Created by Ian Beardsley on 2/9/25.

//

#include <stdio.h>

#include <math.h>

int main(int argc, const char * argv[]) {

float R_p, M_p, R_s, M_s, t_c, M_m, rho_m, rho_p, PlanetDay,

V_p,StarRadius, PlanetRadius, PlanetMass, StarLuminosity, PlanetOrbit,

StarMass, r_p, T_p, p, L_p, KE_p, v_p, T_m,Tmoon, C_m;

float G=6.674E-11, hbarstar, PDCT,Tsquared,T,PlanetYear;

float r_m, R_m, V_m, MoonDensity, part1, part2, part3,v_m, KE_m;

int i;

printf ("What is the radius of the star in solar radii? ");

scanf ("%f", &StarRadius);

printf ("What is the mass of the star in solar masses? ");

scanf ("%f", &StarMass);

printf ("What is the luminosity of the star in solar luminosities? ");

scanf ("%f", &StarLuminosity);

PlanetOrbit=sqrt(StarLuminosity);

r_p=PlanetOrbit*1.496E11;

M_s=1.9891E30*StarMass;

Tsquared=((4*3.14159*3.14159)/(G*M_s))*r_p*r_p*r_p;

T=sqrt(Tsquared);

PlanetYear=T/31557600;

printf("Do you want us to compute the planet radius, 1=yes, 0=no? ");

scanf("%i", &i);

R_s=6.9364E8*StarRadius;

if (i==1)

{

R_s=6.9364E8*StarRadius;

R_p=2*(R_s*R_s)/r_p;

PlanetRadius=R_p/6.378E6;

}

else

{

printf("What is the planet radius in Earth radii?: ");

scanf("%f", &PlanetRadius);

R_p=PlanetRadius*6.378E6;

}

printf("What is the mass of the planet in Earth masses? ");

scanf("%f", &PlanetMass);

M_p=PlanetMass*5.972E24;

printf ("What is the planet day in Earth days? ");

of 63 73

scanf ("%f", &PlanetDay);

T_p=PlanetDay*86400;

printf("That is %f seconds \n", T_p);

{

printf("What is p the pressure gradient exponent of the

protoplanetary disc? ");

scanf("%f", &p);

M_s=1.9891E30*StarMass;

r_m=R_s/1.8;

v_p=sqrt(G*M_s/r_p);

L_p=0.8*3.14159*M_p*(1/T_p)*R_p*R_p;

KE_p=0.5*M_p*v_p*v_p;

hbarstar=L_p/p;

t_c=hbarstar/KE_p;

part1=cbrt(hbarstar/(t_c));

part2=cbrt(1/G);

part3=cbrt(hbarstar/299792458);

M_m=part1*part2*part3;

R_s=StarRadius*6.9634E8;

R_m=R_s*r_m/r_p;

V_m=1.33333*3.14159*R_m*R_m*R_m;

rho_m=(M_m/V_m);

MoonDensity=rho_m*0.001;

V_p=1.33333*3.14159*R_p*R_p*R_p;

rho_p=(M_p/V_p)*0.001;

printf("\n");

printf("Angular Momentum of Planet: %f E33 \n", L_p/

1E33);

printf("\n");

printf("PlanetYear: %f years \n", PlanetYear);

printf("PlanetYear: %f seconds \n", T);

printf("planet orbital velocity: %f m/s \n", v_p);

printf("planet mass: %f E24 kg \n", M_p/1E24);

printf("planet mass: %f Earth masses \n", M_p/5.972E24);

printf("planet radius %f meters \n", R_p);

printf("planet radius: %f Earth Radii \n", PlanetRadius);

printf("planet orbital radius: %f E11 m \n", r_p/1E11);

printf ("planet orbital radius: %f Earth distances \n",

r_p/1.496E11);

printf("planet KE: %f E33 J \n",KE_p/1E33);

printf("planet density: %f g/cm3 \n", rho_p);

printf("\n");

printf("hbarstar: %f E33 Js \n", hbarstar/1E33);

printf("characteristic time: %f seconds\n", t_c);

of 64 73

printf("\n");

printf("Orbital Radius of Moon: %f E8 m \n", r_m/1E8);

printf("Orbital Radius of Moon: %f Moon Distances \n",

r_m/3.84E8);

printf("Radius of Moon: %f E6 m \n", R_m/1E6);

printf("Radius of Moon: %f Moon Radii \n", R_m/1.7374E6);

printf("Mass of Moon: %f E22 kg \n", M_m/1E22);

printf("Mass of Moon %f Moon Masses \n", M_m/

7.347673E22);

printf("density of moon: %f g/cm3 \n", MoonDensity);

printf("\n");

v_m=sqrt(G*M_p/r_m);

KE_m=0.5*M_m*v_m*v_m;

PDCT=(KE_m/KE_p)*(T_p)*(0.91706);

printf("Orbital Velocity of Moon: %f m/s \n", v_m);

printf("PlanetDay Characteristic Time: %f seconds \n",

PDCT);

C_m=2*3.14159*r_m;

T_m=C_m/v_m;

Tmoon=T_m*(1.0/24)*(1.0/60)*(1.0/60);

printf("Lunar Orbital Period: %f seconds \n", T_m);

printf("Lunar Orbital Period: %f days \n", Tmoon);

return 0;

}}

of 65 73

Appendix 4: Defending the Particle Theory

The idea is we find

works with the proton radius what it is, and that of the neutron radius and classical electron

radius. So, the natural constant is 1 second, much in the same way in Newton's Universal Law of

gravity is

We don't say why has the value it has, we measured it and found it works. So it is a Natural

Law. However, I do derive the idea behind it from a hypothesized normal force:

giving

, and so on...

, , ,

And this last one is derived from

Which are correct because when you equate the left side of one to the left side of the other you

get the equation of the radius of a proton is

1 second =

r

i

m

i

⋅

πh

Gc

κ

i

F = G

Mm

r

2

G

F

n

=

h

ct

2

1

, t

1

= 1 second

m

p

=

1

3α

2

⋅

π r

2

p

F

n

G

,

m

e

=

π r

2

eClassical

F

n

G

,

m

n

=

1

3α

2

⋅

π r

2

n

F

n

G

π r

2

p

= AreaCrossSectionProton

1 second =

r

i

m

i

⋅

πh

Gc

κ

i

κ

p

= 1/(3α

2

)

κ

n

= 1/(3α

2

)

κ

e

= 1

r

e

= r

eClassical

(

1

6α

2

4πh

Gc

)

⋅

r

p

m

p

= 1 second

ϕ ⋅

π r

p

α

4

Gm

3

p

1

3

⋅

h

c

= 1 second

of 66 73

Which you can show is correct by looking at Planck energy and mass energy equivalence:

We take the rest energy of the mass of a proton :

The frequency of a proton is

We see at this point we have to set the expression equal to . We explain why this is in a minute

The radius of a proton is then

Something incredible regarding the connection between microscales (the atom's proton) and

macroscales (the solar system) if you want to get very close to modern measurements of the

proton and as well exactly a characteristic time of one second. The radius of a proton is not

constant, but depends of the nature of the experiment, because protons are thought to be a fuzzy

cloud of subatomic particles. We see if we don't use in our equations for protons and the

characteristic time of one second, but the right ratio of terms in the fibonacci sequence that are

approximations to $\phi$, we find that the ratio is 5/8 from the sequence:

If

0, 1, 1, 2, 3, 5, 8, 13,...

is the fibonacci sequence whose successive terms converge on , the golden ratio, then the two

terms that come closest to this are 5/8 = 0.625.

This is a characteristic time from

r

p

= ϕ ⋅

h

cm

p

E = h f

m

p

E = m

p

c

2

f

p

=

m

p

c

2

h

ϕ

m

p

c

2

h

⋅

r

p

c

= ϕ =

m

p

c

h

r

p

m

p

r

p

= ϕ ⋅

h

c

r

p

= ϕ ⋅

h

cm

p

ϕ

r

p

= ϕ

h

cm

p

ϕ =

r

p

m

p

c

h

=

(0.833 × 10

−15

)(1.67262 × 10

−27

)(299,792,458)

6.62607 × 10

−34

= 0.6303866

ϕ

ϕ ⋅

π r

p

α

4

Gm

3

p

1

3

⋅

h

c

= 0.995 seconds

of 67 73

that has a value of

Combining

with

Gives the radius of a proton to be

With this, while we get very close to one second (1.0007 seconds) with the fibonacci ratio of 5/8

we also get something very much in line with the most recent measurement for the radius of a

proton ( ).

5

8

⋅

(352275361)π (0.833 × 10

−15

m)

(6.674 × 10

−11

)(1.67262 × 10

−27

)

3

1

3

⋅

(6.62607 × 10

−34

)

299,792,458

= 1.0007 seconds

5

8

⋅

π r

p

α

4

Gm

3

p

1

3

⋅

h

c

= 1.0007 seconds

(

1

6α

2

4πh

Gc

)

⋅

r

p

m

p

= 1 second

r

p

=

5

8

h

cm

p

r

p

=

5

8

(6.62607 × 10

−34

)

(299,792,458)(1.67262 × 10

−27

)

= 0.8258821 × 10

−15

m

0.831 × 10

−15

m

of 68 73

Appendix 5: Defending the Planetary System Theory

We say the Solar System Planck-type constant is given by"

And, more accurately as (using the ﬁbonacci approximation of 2/3)

Where

=

But we say so because we know it is right from the delocalization time of the Earth which is

given as follows (See Appendix 2 for complete computation)…

ℏ

⊙

= (1secon d )(K E

e

)

ℏ

⊙

= (hC )K E

e

hC = 1secon d

C =

1

3

⋅

1

α

2

c

2

3

⋅

π r

p

G m

3

p

1

3

⋅

18769

299792458

1

3

⋅

2π (0.833E − 15)

(6.67408E − 11)(1.67262E − 27)

3

1.55976565E 33

s

m

kg

3

⋅

s

2

kg

m

3

=

s

m

s

2

kg

2

m

2

=

s

m

⋅

s

kg ⋅ m

=

1

kg

⋅

s

2

m

2

1

C

= kg

m

2

s

2

=

1

2

mv

2

= en erg y

hC = (6.62607E − 34)(1.55976565E 33) = 1.03351secon d s ≈ 1.0secon d s

hC =

(

kg

m

s

2

⋅ m ⋅ s

)

(

1

kg

⋅

s

2

m

2

)

(

kg

m

2

s

)(

1

kg

⋅

s

2

m

2

)

= secon d s

K E

earth

=

1

2

(5.972E 24kg)(30,290m /s)

2

= 2.7396E 33J

ℏ

⊙

= (hC )K E

earth

= (1.03351s)(2.7396E 33J ) = 2.8314E 33J ⋅ s

of 69 73

The Gaussian wavefunction in position space is

It’s Fourier wave decomposition is

We use the Gaussian integral identity (integral of quadratic)

We ﬁnd via the inverse Fourier transform. It is

Substitue :

The solution is standard and is:

Where is the mass of the Moon, and is the orbital radius of the Moon. We

have

ψ (x,0) = Ae

−

x

2

2d

2

ψ (x,0) = Ae

−

x

2

2d

2

=

∫

dp

2π ℏ

ϕ( p)e

i

ℏ

px

∫

∞

−∞

e

−a x

2

+bx

d x =

π

a

e

b

2

4a

ϕ( p)

ϕ( p) =

∫

∞

−∞

d x ψ (x,0)e

−

i

ℏ

px

ψ (x,0)

ϕ( p) = A

∫

∞

−∞

e

−

x

2

2d

2

e

−

i

ℏ

[ px]

d x

|

ψ (x, t)

|

2

=

[

−

x

2

d

2

⋅

1

(1 + t

2

/τ

2

)

]

τ =

m d

2

ℏ

τ =

m

moon

(2r

moon

)

2

ℏ

⊙

m

moon

r

moon

τ = 4

(7.34767E 22kg)(3.844E8m)

2

2.8314E33J ⋅ s

= 15338227secon ds

of 70 73

Now let’s compute a half a year…

(1/2)(365.25)(24)(60)(60)=15778800 seconds

So we see our delocalization time is very close to the half year over which the Earth and

Moon travel from one position to the opposite side of the Sun. The closeness is

So the equation"

"

Is"

This is the ground state distance described in time by introducing the speed of light c. We see

here one second is the minimal quantum unit. This says the Moon is the metric and doing that for

the direct analogy of energy of an atom in wave solution we ﬁnd that Z the atomic number

becomes the radius of the Sun normalized by the Moon, and that it is described in terms of the

Moon. And we see again that the Planck-type constant for the Solar system works, so it is

consistent across the theory working to better than 99% accuracy giving it orbital energy (Kinetic

energy in an approximately circular orbit):

The Earth as it rotates loses energy to the Moon, so its rotation slows down and the Moon’s orbit

grows. We suggest that the characteristic rotation period of the Earth is about 24 hours because

15338227

15778800

100 = 97.2 %

ℏ

2

⊙

GM

3

m

⋅

1

c

= 1second

ℏ

2

⊙

GM

3

m

⋅

1

c

= 1secon d

λ

moon

=

ℏ

2

⊙

GM

3

m

=

(2.8314E 33)

2

(6.67408E − 11)(7.34763E 22kg)

3

= 3.0281E8m

λ

moon

c

=

3.0281E8m

299,792,458m /s

= 1.010secon ds

λ

moon

c

= 1secon d

E

3

= n

R

⊙

R

m

⋅

G

2

M

2

e

M

3

m

2ℏ

2

⊙

of 71 73

this gives the characteristic time of 1 second if we consider the Moon’s and Earth’s kinetic

energies and the inclination of the Earth’s spin ( ) to it orbital plane in the following

equation:

I should make some quick notes:

We might suggest the Moon is the metric for measuring size, and as well will see distance and mass as

well. This comes to us from the condition for a perfect eclipse of the Sun by the Moon, which is:

The Moon optimizes the conditions for life because it holds the Earth at its tilt to its orbit, preventing

weather extremes, extreme hot and extreme cold, allowing for the Seasons.

In order to apply this to other star systems, we have to be able to predict the radius of the habitable planet,

presumably in the n=3 orbit. I found the answer to be in the Vedic literature of India. They noticed that the

diameter of the Sun is about 108 times the diameter of the Earth and that the average distance from the

Sun to the Earth is about 108 solar diameters, with 108 being a signiﬁcant number in Yoga. So I wrote the

equivalent:

radius of the star. The surprising result I found was, after applying it to the stars of all spectral types

from F through K, with their different radii and luminosities (the luminosities determine , the

distances to the habitable zones), that the radius of the planet always came out about the same, about the

radius of the Earth. This may suggest optimally habitable planets are not just a function of the distance

from the star, which determines their temperature, but are functions of their size and mass probably

because they are good for life chemistry, atmospheric composition, and gravity when they are the size and

mass of the Earth.

In order to get , the distance of the habitable planet from the star, we use the inverse square law for

luminosity of the star. If the Earth is in the habitable zone, and if the star is one hundred times brighter

than the Sun, then by the inverse square law the distance to the habitable zone of the planet is 10 times

that of what the Earth is from the Sun. Thus we have in astronomical units the habitable zone of a star is

given by:

the luminosity of the star, and the luminosity of the Sun. We compute the orbital radius of the

Moon…

θ = 23.5

∘

KE

moon

KE

eart h

(24hours)cos(θ ) ≈ 1second

r

earth

r

moon

=

R

⊙

R

moon

R

planet

= 2

R

2

⋆

r

planet

R

⋆

r

planet

r

planet

r

planet

=

L

⋆

L

⊙

AU

L

⋆

L

⊙

of 72 73

Which works for our Solar System, Ag and Au the relative masses of silver and gold atoms.

r

m

= R

⋆

Ag

Au

= R

⋆

/(1.8) =

6.957E 8m

1.8

= 3.865EE 8m

of 73 73

The Author